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第一个部分 前4题 次小生成树
算法：首先如果生成了最小生成树，那么这些树上的所有的边都进行标记。标记为树边。
接下来进行枚举，枚举任意一条不在MST上的边，如果加入这条边，那么肯定会在这棵树上形成一个环，如果还要维护处树的特点
那么就要在这个环上删去一条边，这样他还是树，删掉的边显然是这条链上权值最大边更可能形成次小生成树。那么就有2中方法可以做。
第一种PRIM在prim时候直接可以做出这个从I到J的链上权值最大的值MAX[i][j];
同时可以用kruskal同样方式标记树边，然后DFS跑出MAX[i][j]
其中prim适合稠密图，kruskal适合稀疏图
2份模板
第一份：prim
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 210; const int INF = 0x3f3f3f3f; bool vis[MAXN]; int cost[MAXN][MAXN]; int lowcost[MAXN]; int pre[MAXN],MAX[MAXN][MAXN]; bool used[MAXN][MAXN]; int prim(int cost[][MAXN],int n) { int ret = 0; memset(vis,false,sizeof(vis)); memset(MAX,0,sizeof(MAX)); memset(used,false,sizeof(used)); vis[1] = true; pre[1] = -1; for (int i = 1 ; i <= n ; i++) { lowcost[i] = cost[1][i]; pre[i] = 1; } lowcost[1] = 0; pre[1] = -1; for (int i = 2 ; i <= n ; i++) { int mincost = INF; int pos = -1; for (int j = 1 ; j <= n ; j++) { if (!vis[j] && lowcost[j] < mincost) { mincost = lowcost[j]; pos = j; } } if (pos == -1) return -1; ret += mincost; vis[pos] = true; used[pos][pre[pos]] = used[pre[pos]][pos] = true; for (int j = 1; j <= n ; j++) { if (vis[j] && j != pos) MAX[j][pos] = MAX[pos][j] = max(MAX[j][pre[pos]],lowcost[pos]); if (!vis[j] && lowcost[j] > cost[pos][j]) { lowcost[j] = cost[pos][j]; pre[j] = pos; } } } return ret; } int calcusmst(int cost[][MAXN],int n,int mst) { int MIN = INF; for (int i = 1 ; i <= n ; i++) { for (int j = i + 1 ; j <= n ; j++) { if (cost[i][j] < INF && !used[i][j]) { if (MIN > mst + cost[i][j] - MAX[i][j]) MIN = mst + cost[i][j] - MAX[i][j]; } } } return MIN; } int main() { // freopen("sample.txt","r",stdin); int T; scanf("%d",&T); while (T--) { int N,M; scanf("%d%d",&N,&M); for (int i = 0 ; i <= N ; i++) for (int j = 0 ; j <= N ; j++) cost[i][j] = i == j ? 0 : INF; while (M--) { int u,v,w; scanf("%d%d%d",&u,&v,&w); cost[u][v] = w; cost[v][u] = w; } int ret = prim(cost,N); int smst = calcusmst(cost,N,ret); printf("%d %d\n",ret,smst); } return 0; }
第二份：kruskal
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 110; const int MAXM = MAXN * MAXN; const int INF = 0x3f3f3f3f; struct node { int u,v,w; bool key; friend bool operator < (const node &a,const node &b) { return a.w < b.w; } }edge[MAXM]; int fa[MAXN]; int MAX[MAXN][MAXN]; int Find(int x) {return x == fa[x] ? x : fa[x] = Find(fa[x]);} int N,M; struct Edge { int u,v,w; }; vector<Edge>G[MAXN]; vector<int>res; int kruskal(int N) { sort(edge,edge + M); int ret = 0,cnt = 0; for (int i = 0 ; i < M ; i++) { int fu = Find(edge[i].u); int fv = Find(edge[i].v); if (fu != fv) { fa[fv] = fu; cnt++; ret += edge[i].w; edge[i].key = true; } if(cnt >= N - 1) break; } return ret; } void dfs(int u,int fa,int facost) { for (int i = 0 ; i < (int)res.size() ; i++) { int x = res[i]; MAX[x][u] = MAX[u][x] = max(MAX[fa][x],facost); } res.push_back(u); for (int i = 0 ; i < (int)G[u].size(); i++) { int v = G[u][i].v; if (v != fa) { dfs(v,u,G[u][i].w); } } } int main() { // freopen("sample.txt","r",stdin); int T; scanf("%d",&T); while (T--) { scanf("%d%d",&N,&M); for (int i = 0 ; i <= N ; i++) fa[i] = i; for (int i = 0 ; i < M ; i++) { scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w); edge[i].key = false; } for (int i = 0 ; i <= N ; i++) G[i].clear(); int mst = kruskal(N); for (int i = 0 ; i < M ; i++) { if (edge[i].key == false) continue; Edge tmp; tmp.u = edge[i].u; tmp.v = edge[i].v; tmp.w = edge[i].w; G[edge[i].u].push_back(tmp); tmp.v = edge[i].u; tmp.u = edge[i].v; G[edge[i].v].push_back(tmp); } memset(MAX,0,sizeof(MAX)); res.clear(); int smst = INF; dfs(1,-1,0); for (int i = 0 ; i < M ; i++) { if (!edge[i].key) smst = min(smst,mst + edge[i].w - MAX[edge[i].u][edge[i].v]); } printf("%d %d\n",mst,smst); } return 0; }
POJ 1679 The Unique MST
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 210; const int INF = 0x3f3f3f3f; bool vis[MAXN]; int cost[MAXN][MAXN]; int lowcost[MAXN]; int pre[MAXN],MAX[MAXN][MAXN]; bool used[MAXN][MAXN]; int prim(int cost[][MAXN],int n) { int ret = 0; memset(vis,false,sizeof(vis)); memset(MAX,0,sizeof(MAX)); memset(used,false,sizeof(used)); vis[1] = true; pre[1] = -1; for (int i = 1 ; i <= n ; i++) { lowcost[i] = cost[1][i]; pre[i] = 1; } lowcost[1] = 0; pre[1] = -1; for (int i = 2 ; i <= n ; i++) { int mincost = INF; int pos = -1; for (int j = 1 ; j <= n ; j++) { if (!vis[j] && lowcost[j] < mincost) { mincost = lowcost[j]; pos = j; } } if (pos == -1) return -1; ret += mincost; vis[pos] = true; used[pos][pre[pos]] = used[pre[pos]][pos] = true; for (int j = 1; j <= n ; j++) { if (vis[j]) MAX[j][pos] = MAX[pos][j] = max(MAX[j][pre[pos]],lowcost[pos]); if (!vis[j] && lowcost[j] > cost[pos][j]) { lowcost[j] = cost[pos][j]; pre[j] = pos; } } } return ret; } int calcusmst(int cost[][MAXN],int n,int ret) { int MIN = INF; for (int i = 1 ; i <= n ; i++) { for (int j = i + 1 ; j <= n ; j++) { if (cost[i][j] < INF && !used[i][j]) MIN = min(MIN,ret + cost[i][j] - MAX[i][j]); } } if (MIN == INF) return -1; return MIN; } int main() { int T,N,M; scanf("%d",&T); while (T--) { scanf("%d%d",&N,&M); for (int i = 0 ; i <= N ; i++) for (int j = 0 ; j <= N ; j++) cost[i][j] = i == j ? 0 : INF; while (M--) { int u,v,w; scanf("%d%d%d",&u,&v,&w); cost[u][v] = cost[v][u] = w; } int ret = prim(cost,N); if (ret == -1) { printf("Not Unique!\n"); continue; } if (ret == calcusmst(cost,N,ret)) printf("Not Unique!\n"); else printf("%d\n",ret); } return 0; }
HDU 4081 Qin Shi Huang‘s National Road System
这个题的关键边可以是树边可以非树边主要需要MAX[i][j]数组
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const double INF = 1e14; const int MAXN = 1010; struct point { int x,y; int people; }src[MAXN]; int N; double edge[MAXN][MAXN]; int nearvex[MAXN]; double lowcost[MAXN]; double sum; bool used[MAXN][MAXN]; bool vis[MAXN]; double MAX[MAXN][MAXN]; void prim(int v0) { sum = 0; memset(used,false,sizeof(used)); memset(vis,false,sizeof(vis)); memset(MAX,0,sizeof(MAX)); for (int i = 1 ; i <= N ; i++) { lowcost[i] = edge[v0][i]; nearvex[i] = v0; } vis[v0] = true; for (int i = 1 ; i <= N ; i++) { double MIN = INF; int pos = -1; for (int j = 1 ; j <= N ; j++) { if (!vis[j] && lowcost[j] < MIN) { MIN = lowcost[j]; pos = j; } } if (pos != -1) { sum += MIN; used[pos][nearvex[pos]] = used[nearvex[pos]][pos] = true; vis[pos] = true; for (int k = 1 ; k <= N ; k++) { if (vis[k] && k != pos) { MAX[pos][k] = MAX[k][pos] = max(lowcost[pos],MAX[k][nearvex[pos]]); } else if (!vis[k] && edge[pos][k] < lowcost[k]) { lowcost[k] = edge[pos][k]; nearvex[k] = pos; } } } } } int main() { int T; scanf("%d",&T); while (T--) { scanf("%d",&N); for (int i = 1 ; i <= N ; i++)scanf("%d%d%d",&src[i].x,&src[i].y,&src[i].people); for (int i = 1 ; i <= N ; i++) { edge[i][i] = 0.0; for (int j = i + 1 ; j <= N ; j++) { double dist = sqrt((src[j].x - src[i].x) * (src[j].x - src[i].x) + (src[j].y - src[i].y) * (src[j].y - src[i].y)); edge[i][j] = edge[j][i] = dist; } } prim(1); double ret = -1.0; for (int i = 1 ; i <= N ; i++) { for (int j = 1 ; j <= N ; j++) { if (i == j) continue; if (used[i][j]) { double tmp = 1.0 * (src[i].people + src[j].people) / (sum - edge[i][j]); ret = max(tmp,ret); } else { double tmp = 1.0 * (src[i].people + src[j].people) / (sum - MAX[i][j]); ret = max(ret,tmp); } } } printf("%.2lf\n",ret); } return 0; }
HDU 10600 ACM Contest and Blackout
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 110; const int MAXM = MAXN * MAXN; const int INF = 0x3f3f3f3f; struct node { int u,v,w; bool key; friend bool operator < (const node &a,const node &b) { return a.w < b.w; } }edge[MAXM]; int fa[MAXN]; int MAX[MAXN][MAXN]; int Find(int x) {return x == fa[x] ? x : fa[x] = Find(fa[x]);} int N,M; struct Edge { int u,v,w; }; vector<Edge>G[MAXN]; vector<int>res; int kruskal(int N) { sort(edge,edge + M); int ret = 0,cnt = 0; for (int i = 0 ; i < M ; i++) { int fu = Find(edge[i].u); int fv = Find(edge[i].v); if (fu != fv) { fa[fv] = fu; cnt++; ret += edge[i].w; edge[i].key = true; } if(cnt >= N - 1) break; } return ret; } void dfs(int u,int fa,int facost) { for (int i = 0 ; i < (int)res.size() ; i++) { int x = res[i]; MAX[x][u] = MAX[u][x] = max(MAX[fa][x],facost); } res.push_back(u); for (int i = 0 ; i < (int)G[u].size(); i++) { int v = G[u][i].v; if (v != fa) { dfs(v,u,G[u][i].w); } } } int main() { // freopen("sample.txt","r",stdin); int T; scanf("%d",&T); while (T--) { scanf("%d%d",&N,&M); for (int i = 0 ; i <= N ; i++) fa[i] = i; for (int i = 0 ; i < M ; i++) { scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w); edge[i].key = false; } for (int i = 0 ; i <= N ; i++) G[i].clear(); int mst = kruskal(N); for (int i = 0 ; i < M ; i++) { if (edge[i].key == false) continue; Edge tmp; tmp.u = edge[i].u; tmp.v = edge[i].v; tmp.w = edge[i].w; G[edge[i].u].push_back(tmp); tmp.v = edge[i].u; tmp.u = edge[i].v; G[edge[i].v].push_back(tmp); } memset(MAX,0,sizeof(MAX)); res.clear(); int smst = INF; dfs(1,-1,0); for (int i = 0 ; i < M ; i++) { if (!edge[i].key) smst = min(smst,mst + edge[i].w - MAX[edge[i].u][edge[i].v]); } printf("%d %d\n",mst,smst); } return 0; }
UVA 10462 Is There A Second Way Left?
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 110; const int MAXM = 410; const int INF = 0x3f3f3f3f; struct node { int u,v,w; bool key; friend bool operator < (const node &a,const node &b) { return a.w < b.w; } }edge[MAXM]; int fa[MAXN]; int Find(int x) {return x == fa[x] ? x : fa[x] = Find(fa[x]);} int MAX[MAXN][MAXN]; int N,M; vector<node>G[MAXN]; vector<int>res; int kruskal(int N) { for (int i = 0 ; i <= N ; i++) fa[i] = i; sort(edge,edge + M); int cnt = 0,ret = 0; bool flag = false; for (int i = 0 ; i < M ; i++) { int fu = Find(edge[i].u); int fv = Find(edge[i].v); if (fv != fu) { fa[fv] = fu; cnt++; ret += edge[i].w; edge[i].key = true; } if (cnt >= N - 1) { flag = true; break; } } if (flag) return ret; return -1; } void dfs(int u,int fa,int precost) { for (int i = 0 ; i < (int)res.size() ; i++) { int x = res[i]; MAX[x][u] = MAX[u][x] = max(MAX[fa][x],precost); } res.push_back(u); for (int i = 0 ; i < (int)G[u].size() ; i++) { int v = G[u][i].v; if (v == fa) continue; dfs(v,u,G[u][i].w); } } int calcusmst(int N,int mst) { int ret = INF; for (int i = 0 ; i < M ; i++) { if (edge[i].key == false) { ret = min(ret,mst + edge[i].w - MAX[edge[i].u][edge[i].v]); } } return ret == INF ? -1 : ret; } int main() { // freopen("sample.txt","r",stdin); int T,kase = 1; scanf("%d",&T); while (T--) { scanf("%d%d",&N,&M); for (int i = 0 ; i < M ; i++) { scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w); edge[i].key = false; } if (N == 1) { printf("Case #%d : No second way\n",kase++); continue; } for (int i = 0 ; i <= N ; i++) G[i].clear(); int mst = kruskal(N); if (mst == -1) { printf("Case #%d : No way\n",kase++); continue; } for (int i = 0 ; i < M ; i++) { if (edge[i].key == false) continue; int u = edge[i].u; int v = edge[i].v; node tmp; tmp.u = u; tmp.v = v; tmp.w = edge[i].w; G[u].push_back(tmp); swap(tmp.v,tmp.u); G[v].push_back(tmp); } res.clear(); memset(MAX,0,sizeof(MAX)); dfs(1,-1,0); int smst = calcusmst(N,mst); if (smst == -1) { printf("Case #%d : No second way\n",kase++); } else printf("Case #%d : %d\n",kase++,smst); } return 0; }
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下面是最小树形图。
一些很好的资料
http://alanyume.com/669.html
假设一个有向图有3个顶点
1->2 8,
1->3 8,
2->3 4,
3->2 3
四条边，由于1->2和1->3是相等的，所以prim因为循环顺序的原因，
最后结果构造出来的树是1->2,2->3但是答案应该是1->3->2，无向图就没有这种问题。
模板
const int MAXN = 1010; const int MAXM = MAXN * MAXN; const int INF = 0x3f3f3f3f; typedef int type; struct Edge { int u,v; type w; }edge[MAXM]; int pre[MAXN], id[MAXN], vis[MAXN], N,M; type in[MAXN]; type Zhuliu(int root, int V, int E) { type ret = 0; while(true) { //1.找最小入边 for(int i = 0; i < V; i++) in[i] = INF; for(int i = 0; i < E; i++) { int u = edge[i].u; int v = edge[i].v; if(edge[i].w < in[v] && u != v) {pre[v] = u; in[v] = edge[i].w;} } for(int i = 0; i < V; i++) { if(i == root) continue; if(in[i] == INF) return -1;//除了跟以外有点没有入边,则根无法到达它 } //2.找环 int cnt = 0; memset(id, -1, sizeof(id)); memset(vis, -1, sizeof(vis)); in[root] = 0; for(int i = 0; i < V; i++) //标记每个环 { ret += in[i]; int v = i; while(vis[v] != i && id[v] == -1 && v != root) //每个点寻找其前序点，要么最终寻找至根部，要么找到一个环 { vis[v] = i; v = pre[v]; } if(v != root && id[v] == -1)//缩点 { for(int u = pre[v]; u != v; u = pre[u]) id[u] = cnt; id[v] = cnt++; } } if(cnt == 0) break; //无环 则break for(int i = 0; i < V; i++) if(id[i] == -1) id[i] = cnt++; //3.建立新图 for(int i = 0; i < E; i++) { int u = edge[i].u; int v = edge[i].v; edge[i].u = id[u]; edge[i].v = id[v]; if(id[u] != id[v]) edge[i].w -= in[v]; } V = cnt; root = id[root]; } return ret; }
POJ 3164 Command Network
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 110; const int MAXM = MAXN * MAXN; const double INF = 1e15; typedef double type; struct point { double x,y; }src[MAXN]; struct Edge { int u,v; type w; }edge[MAXM]; int pre[MAXN], id[MAXN], vis[MAXN], N,M; type in[MAXN]; double dis(point a, point b) { return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); } type Zhuliu(int root, int V, int E) { type ret = 0; while(true) { //1.找最小入边 for(int i = 0; i < V; i++) in[i] = INF; for(int i = 0; i < E; i++) { int u = edge[i].u; int v = edge[i].v; if(edge[i].w < in[v] && u != v) {pre[v] = u; in[v] = edge[i].w;} } for(int i = 0; i < V; i++) { if(i == root) continue; if(in[i] == INF) return -1;//除了跟以外有点没有入边,则根无法到达它 } //2.找环 int cnt = 0; memset(id, -1, sizeof(id)); memset(vis, -1, sizeof(vis)); in[root] = 0; for(int i = 0; i < V; i++) //标记每个环 { ret += in[i]; int v = i; while(vis[v] != i && id[v] == -1 && v != root) //每个点寻找其前序点，要么最终寻找至根部，要么找到一个环 { vis[v] = i; v = pre[v]; } if(v != root && id[v] == -1)//缩点 { for(int u = pre[v]; u != v; u = pre[u]) id[u] = cnt; id[v] = cnt++; } } if(cnt == 0) break; //无环 则break for(int i = 0; i < V; i++) if(id[i] == -1) id[i] = cnt++; //3.建立新图 for(int i = 0; i < E; i++) { int u = edge[i].u; int v = edge[i].v; edge[i].u = id[u]; edge[i].v = id[v]; if(id[u] != id[v]) edge[i].w -= in[v]; } V = cnt; root = id[root]; } return ret; } int main() { // freopen("sample.txt","r",stdin); while (scanf("%d%d",&N,&M) != EOF) { for (int i = 0 ; i < N ; i++) scanf("%lf%lf",&src[i].x,&src[i].y); for (int i = 0 ; i < M ; i++) { int u,v; scanf("%d%d",&u,&v); u--;v--; edge[i].u = u; edge[i].v = v; if (edge[i].u == edge[i].v) edge[i].w = INF; else edge[i].w = dis(src[u],src[v]); } type ans = Zhuliu(0,N,M); if (ans == -1) puts("poor snoopy"); else printf("%.2f\n",ans); } return 0; }
UVA 11183 Teen Girl Squad
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 1010; const int MAXM = MAXN * MAXN; const int INF = 0x3f3f3f3f; typedef int type; struct Edge { int u,v; type w; }edge[MAXM]; int pre[MAXN], id[MAXN], vis[MAXN], N,M; type in[MAXN]; type Zhuliu(int root, int V, int E) { type ret = 0; while(true) { //1.找最小入边 for(int i = 0; i < V; i++) in[i] = INF; for(int i = 0; i < E; i++) { int u = edge[i].u; int v = edge[i].v; if(edge[i].w < in[v] && u != v) {pre[v] = u; in[v] = edge[i].w;} } for(int i = 0; i < V; i++) { if(i == root) continue; if(in[i] == INF) return -1;//除了跟以外有点没有入边,则根无法到达它 } //2.找环 int cnt = 0; memset(id, -1, sizeof(id)); memset(vis, -1, sizeof(vis)); in[root] = 0; for(int i = 0; i < V; i++) //标记每个环 { ret += in[i]; int v = i; while(vis[v] != i && id[v] == -1 && v != root) //每个点寻找其前序点，要么最终寻找至根部，要么找到一个环 { vis[v] = i; v = pre[v]; } if(v != root && id[v] == -1)//缩点 { for(int u = pre[v]; u != v; u = pre[u]) id[u] = cnt; id[v] = cnt++; } } if(cnt == 0) break; //无环 则break for(int i = 0; i < V; i++) if(id[i] == -1) id[i] = cnt++; //3.建立新图 for(int i = 0; i < E; i++) { int u = edge[i].u; int v = edge[i].v; edge[i].u = id[u]; edge[i].v = id[v]; if(id[u] != id[v]) edge[i].w -= in[v]; } V = cnt; root = id[root]; } return ret; } int main() { // freopen("sample.txt","r",stdin); int T,kase = 1; scanf("%d",&T); while (T--) { int N,M; scanf("%d%d",&N,&M); for (int i = 0 ; i < M ; i++) { scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w); } type ret = Zhuliu(0,N,M); if (ret == -1) printf("Case #%d: Possums!\n",kase++); else printf("Case #%d: %d\n",kase++,ret); } return 0; }
HDU 2121 Ice_cream’s world II
非固定根的最小树形图，这里添加虚拟根，连接各个顶点。权值要足够大，这里权值就设置所有输入边权值和+1。这个值为sum
如果最后最小树形图代价大于等于2*sum就说明利用2条虚拟边作为树边这种情况就是最小树形图不存在
另外注意还要输出最优的那个根，这里要根据边的关系来做。因为在zhuliu算法中每次缩点后都要给所有点从新编号，对于这个题输出答案就很麻烦，
所以就从边上下手；
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 //http://alanyume.com/669.html /* 假设一个有向图有3个顶点 1->2 8, 1->3 8, 2->3 4, 3->2 3 四条边，由于1->2和1->3是相等的，所以prim因为循环顺序的原因， 最后结果构造出来的树是1->2,2->3但是答案应该是1->3->2，无向图就没有这种问题。 using namespace std; */ int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 1010; const int MAXM = MAXN * MAXN; const int INF = 0x3f3f3f3f; typedef int type; struct Edge { int u,v; type w; }edge[MAXM]; int pre[MAXN], id[MAXN], vis[MAXN], N,M,pos; type in[MAXN]; type Zhuliu(int root, int V, int E) { type ret = 0; while(true) { //1.找最小入边 for(int i = 0; i < V; i++) in[i] = INF; for(int i = 0; i < E; i++) { int u = edge[i].u; int v = edge[i].v; if(edge[i].w < in[v] && u != v) {pre[v] = u; in[v] = edge[i].w; if (u == root) pos = i;} // 因为在缩点的时候我们需要给每一个点进行从新编号，这样对于我们是很尴尬的，于是我们只能从边上下手咯， // 我们在每次对点进行从新编号的时候记录下虚拟节点的出边编号 } for(int i = 0; i < V; i++) { if(i == root) continue; if(in[i] == INF) return -1;//除了跟以外有点没有入边,则根无法到达它 } //2.找环 int cnt = 0; memset(id, -1, sizeof(id)); memset(vis, -1, sizeof(vis)); in[root] = 0; for(int i = 0; i < V; i++) //标记每个环 { ret += in[i]; int v = i; while(vis[v] != i && id[v] == -1 && v != root) //每个点寻找其前序点，要么最终寻找至根部，要么找到一个环 { vis[v] = i; v = pre[v]; } if(v != root && id[v] == -1)//缩点 { for(int u = pre[v]; u != v; u = pre[u]) id[u] = cnt; id[v] = cnt++; } } if(cnt == 0) break; //无环 则break for(int i = 0; i < V; i++) if(id[i] == -1) id[i] = cnt++; //3.建立新图 for(int i = 0; i < E; i++) { int u = edge[i].u; int v = edge[i].v; edge[i].u = id[u]; edge[i].v = id[v]; if(id[u] != id[v]) edge[i].w -= in[v]; } V = cnt; root = id[root]; } return ret; } int main() { // freopen("sample.txt","r",stdin); while (scanf("%d%d",&N,&M) != EOF) { type sum = 0; int tot = 0; for (int i = 0 ; i < M ; i++) { scanf("%d%d%d",&edge[tot].u,&edge[tot].v,&edge[tot].w); edge[tot].u++; edge[tot].v++; sum += edge[tot].w; tot++; } sum++; for (int i = 1 ; i <= N ; i++) { edge[tot].u = 0; edge[tot].v = i; edge[tot].w = sum; tot++; } // for (int i = 0 ; i < tot ; i++) printf("%d %d %d\n",edge[i].u,edge[i].v,edge[i].w); type ret = Zhuliu(0,N + 1,tot); if (ret == -1 || ret >= 2 * sum) printf("impossible\n"); else { int ans; for (int i = 1 ; i <= N ; i++) if (pre[i] == 0) { ans = i; break; } printf("%d %d\n",ret - sum,pos - M); } puts(""); } return 0; }
HDU 4009 Transfer water
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 1010; const int MAXM = MAXN * MAXN; const int INF = 0x3f3f3f3f; struct Edge { int u,v,next; int cost; }edge[MAXM]; int pre[MAXN],id[MAXN],visit[MAXN],in[MAXN]; int zhuliu(int root,int n,int m,Edge edge[]) { int res = 0,u ,v; while (true) { for (int i = 0 ; i < n ; i++) in[i] = INF; for (int i = 0 ; i < m ; i++) { if (edge[i].u != edge[i].v && edge[i].cost < in[edge[i].v]) { pre[edge[i].v] = edge[i].u; in[edge[i].v] = edge[i].cost; } } for (int i = 0 ; i < n ; i++) if (i != root && in[i] == INF) return -1; //no tree; int tn = 0; memset(id,-1,sizeof(id)); memset(visit,-1,sizeof(visit)); in[root] = 0; for (int i = 0 ; i < n ; i++) { res += in[i]; v = i; while (visit[v] != i && id[v] == -1 && v != root) { visit[v] = i; v = pre[v]; } if (v != root && id[v] == -1) { for (int u = pre[v] ; u != v ; u = pre[u]) id[u] = tn; id[v] = tn++; } } if (tn == 0) break; for (int i = 0 ; i < n ; i++) if (id[i] == -1) id[i] = tn++; for (int i = 0 ; i < m ;) { v = edge[i].v; edge[i].u = id[edge[i].u]; edge[i].v = id[edge[i].v]; if (edge[i].u != edge[i].v) edge[i++].cost -= in[v]; else swap(edge[i],edge[--m]); } n = tn; root = id[root]; } return res; } int N,X,Y,Z; struct node { int a,b,c; }src[MAXN]; int main() { while (scanf("%d%d%d%d",&N,&X,&Y,&Z) != EOF) { if (N == 0 && X == 0 && Y == 0 && Z == 0) break; int L = 0; for (int i = 1 ; i <= N ; i++) { scanf("%d%d%d",&src[i].a,&src[i].b,&src[i].c); edge[L].u = 0; edge[L].v = i; edge[L].cost = src[i].c * X; L++; } for (int i = 1 ; i <= N ; i++) { int cnt ; scanf("%d",&cnt); for (int j = 0 ; j < cnt ; j++) { int x; scanf("%d",&x); if (x == i) continue; edge[L].u = i; edge[L].v = x; if (src[i].c >= src[x].c) edge[L++].cost = Y * (abs(src[x].a - src[i].a) + abs(src[x].b - src[i].b) + abs(src[x].c - src[i].c)); else edge[L++].cost = Z + Y * (abs(src[x].a - src[i].a) + abs(src[x].b - src[i].b) + abs(src[x].c - src[i].c)); } } int ret = zhuliu(0,N + 1,L,edge); if (ret == -1) puts("poor XiaoA\n"); else printf("%d\n",ret); } return 0; }
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生成树计数
Matrix-Tree定理(Kirchhoff矩阵-树定理)。Matrix-Tree定理是解决生成树计数问题最有力的武器之一。它首先于1847年被Kirchhoff证明。在介绍定理之前，我们首先明确几个概念：
1、G的度数矩阵D[G]是一个n*n的矩阵，并且满足：当i≠j时,dij=0；当i=j时，dij等于vi的度数。
2、G的邻接矩阵A[G]也是一个n*n的矩阵， 并且满足：如果vi、vj之间有边直接相连，则aij=1，否则为0。
我们定义G的Kirchhoff矩阵(也称为拉普拉斯算子)C[G]为C[G]=D[G]-A[G]，则Matrix-Tree定理可以描述为：G的所有不同的生成树的个数等于其Kirchhoff矩阵C[G]任何一个n-1阶主子式的行列式的绝对值。所谓n-1阶主子式，就是对于r(1≤r≤n)，将C[G]的第r行、第r列同时去掉后得到的新矩阵，用Cr[G]表示。
模板：
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 20; const int dx[] = {0,0,-1,1}; const int dy[] = {1,-1,0,0}; int N,M; typedef long long type; const long long MOD = 1e9; int id[20][20]; char G[20][20]; int cas; struct Matrix { type mat[MAXN][MAXN]; void init() { memset(mat,0,sizeof(mat)); } type det(int n) { int sign = 0; type ret = 1; for (int i = 0 ; i < n ; i++) { for (int j = i + 1 ; j < n ; j++) { while(mat[j][i]) { type tmp = mat[i][i] / mat[j][i]; for (int k = i ; k < n ; k++) { mat[i][k] = (mat[i][k] - tmp * mat[j][k]); // %MOD swap(mat[i][k],mat[j][k]); } sign++; } } if (mat[i][i] == 0) return 0; ret = (ret * mat[i][i]);// % MOD } if (sign & 1) ret = -ret; return ret;//(ret % MOD + MOD) % MOD; } }slover; int g[MAXN][MAXN]; int main() { int T; scanf("%d",&T); while (T--) { int N,M; scanf("%d%d",&N,&M); memset(g,0,sizeof(g)); while (M--) { int u,v; scanf("%d%d",&u,&v); u--; v--; g[u][v] = g[v][u] = 1; } slover.init(); for (int i = 0 ; i < N ; i++) { for (int j = 0 ; j < N ; j++) { if (i != j && g[i][j]) { slover.mat[i][j] = -1; slover.mat[i][i]++; } } } printf("%lld\n",slover.det(N - 1)); } return 0; }
UVA 10766 Organising the Organisation
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const double eps = 1e-8; const int MAXN = 110; int sgn(long double x) { if (fabs(x) < eps) return 0; if (x < 0) return -1; else return 1; } long double b[MAXN][MAXN]; long double det(long double a[][MAXN],int n) { int i, j, k, sign = 0; long double ret = 1; for(i = 0 ; i < n ; i++) for(j = 0 ; j < n ; j++) b[i][j] = a[i][j]; for(i = 0;i < n;i++) { if(sgn(b[i][i]) == 0) { for(j = i + 1 ; j < n ; j++) if(sgn(b[j][i]) != 0) break; if(j == n)return 0; for(k = i ; k < n ; k++) swap(b[i][k],b[j][k]); sign++; } ret *= b[i][i]; for(k = i + 1 ; k < n ; k++) b[i][k] /= b[i][i]; for(j = i + 1 ; j < n ; j++) for(k = i + 1 ; k < n ; k++) b[j][k] -= b[j][i]*b[i][k]; } if(sign & 1)ret = -ret; return ret; } long double a[MAXN][MAXN]; int g[MAXN][MAXN]; int main() { int n,m,k; while (scanf("%d%d%d",&n,&m,&k) != EOF) { memset(g,0,sizeof(g)); while (m--) { int u,v; scanf("%d%d",&u,&v); u--;v--; g[u][v] = g[v][u] = 1; } memset(a,0,sizeof(a)); for (int i = 0 ; i < n ; i++) for (int j = 0 ; j < n ; j++) { if (i != j && g[i][j] == 0) { a[i][i]++; a[i][j] = -1; } } double ans = det(a,n - 1); printf("%.0lf\n",ans); } return 0; }
SPOJ DETER3 DETER3 - Find The Determinant III
计算行列式
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 250; int N; typedef long long type; type MOD; struct Matrix { type mat[MAXN][MAXN]; void init() { memset(mat,0,sizeof(mat)); } type det(int n) { int sign = 0; type ret = 1; for (int i = 0 ; i < n ; i++) { for (int j = i + 1 ; j < n ; j++) { while(mat[j][i]) { type tmp = mat[i][i] / mat[j][i]; for (int k = i ; k < n ; k++) { mat[i][k] = (mat[i][k] - tmp * mat[j][k]) % MOD; swap(mat[i][k],mat[j][k]); } sign++; } } if (mat[i][i] == 0) return 0; ret = (ret * mat[i][i]) % MOD; } if (sign & 1) ret = -ret; return (ret % MOD + MOD) % MOD; } }slover; int main() { while (scanf("%d%lld",&N,&MOD) != EOF) { for (int i = 0 ; i < N ; i++) for (int j = 0 ; j < N ; j++) scanf("%lld",&slover.mat[i][j]); printf("%lld\n",slover.det(N)); } return 0; }
URAL 1627 Join
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 150; const int dx[] = {0,0,-1,1}; const int dy[] = {1,-1,0,0}; int N,M; typedef long long type; const long long MOD = 1e9; int id[20][20]; char G[20][20]; int cas; struct Matrix { type mat[MAXN][MAXN]; void init() { memset(mat,0,sizeof(mat)); } type det(int n) { int sign = 0; type ret = 1; for (int i = 0 ; i < n ; i++) { for (int j = i + 1 ; j < n ; j++) { while(mat[j][i]) { type tmp = mat[i][i] / mat[j][i]; for (int k = i ; k < n ; k++) { mat[i][k] = (mat[i][k] - tmp * mat[j][k]) % MOD; swap(mat[i][k],mat[j][k]); } sign++; } } if (mat[i][i] == 0) return 0; ret = (ret * mat[i][i]) % MOD; } if (sign & 1) ret = -ret; return (ret % MOD + MOD) % MOD; } }slover; int g[MAXN][MAXN]; int main() { while (scanf("%d%d",&N,&M) != EOF) { cas = 0; for (int i = 0 ; i < N ; i++) scanf("%s",G[i]); memset(id,-1,sizeof(id)); for (int i = 0 ; i < N ; i++) { for (int j = 0 ; j < M ; j++) { if (G[i][j] == ‘.‘) id[i][j] = cas++; } } memset(g,0,sizeof(g)); for (int i = 0 ; i < N ; i++) { for (int j = 0 ; j < M ; j++) { if (id[i][j] != -1) { int l = id[i][j]; for (int d = 0 ; d < 4 ; d++) { int nx = i + dx[d]; int ny = j + dy[d]; if (nx >= 0 && nx < N && ny >= 0 && ny < M && id[nx][ny] != -1) { int r = id[nx][ny]; g[l][r] = g[r][l] = 1; } } } } } slover.init(); for (int i = 0 ; i < cas ; i++) { for (int j = 0 ; j < cas ; j++) { if (i != j && g[i][j]) { slover.mat[i][j] = -1; slover.mat[i][i]++; } } } printf("%I64d\n",slover.det(cas - 1)); } return 0; }
HDU 4305 Lighting
这个题处理很麻烦。计算几何完全不会。。看了bin神代码写的。不过思路很简单。
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MOD = 10007; const int MAXN = 350; int N,R; typedef int type; void ext_gcd(type a,type b,type &d, type &x,type &y) { if (b == 0) {d = a ; x = 1 ; y = 0;} else {ext_gcd(b,a % b,d,y,x); y -= x * (a / b);} } type inv(type a,type n) { type d,x,y; ext_gcd(a,n,d,x,y); return d == 1 ? (x + n) % n : -1; } struct Matrix { type mat[MAXN][MAXN]; void init() { memset(mat,0,sizeof(mat)); } type det(int n) { for (int i = 0 ; i < n ; i++) for (int j = 0 ; j < n ; j++) mat[i][j] = (mat[i][j] % MOD + MOD) % MOD; type ret = 1; for (int i = 0 ; i < n ; i++) { for (int j = i ; j < n ; j++) { if (mat[j][i] != 0) { for (int k = i ; k < n ; k++) swap(mat[i][k],mat[j][k]); if (i != j) ret = (-ret + MOD) % MOD; break; } } if (mat[i][i] == 0) { ret = -1; break; } for (int j = i + 1 ; j < n ; j++) { // int mut = (mat[j][i] * INV[mat[i][i]]) % MOD; int mut = (mat[j][i] * inv(mat[i][i],MOD)) % MOD; for (int k = i ; k < n ; k++) mat[j][k] = (mat[j][k] - (mat[i][k] * mut) % MOD + MOD) % MOD; } ret = (ret * mat[i][i]) % MOD; } return ret; } }slover; struct point { int x,y; point (int tx = 0,int ty = 0) { x = tx; y = ty; } point operator - (const point &rhs) const { return point(x - rhs.x,y - rhs.y); } int operator ^ (const point &rhs) const { return x * rhs.y - y * rhs.x; } }src[MAXN]; struct Line { point s,e; Line(){} Line(point a,point b) { s = a; e = b; } }; bool onseg(point p,Line L) { return ((L.s - p) ^ (L.e - p)) == 0 && (p.x - L.s.x) * (p.x - L.e.x) <= 0 && (p.y - L.s.y) * (p.y - L.e.y) <= 0; } type dist(point a,point b) { return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y); } bool judge(int a,int b) { if (dist(src[a],src[b]) > R * R) return false; for (int i = 0 ; i < N ; i++) { if (i == a || i == b) continue; if (onseg(src[i],Line(src[a],src[b]))) return false; } return true; } int g[MAXN][MAXN]; int main() { int T; scanf("%d",&T); while (T--) { scanf("%d%d",&N,&R); for (int i = 0 ; i < N ; i++) scanf("%d%d",&src[i].x,&src[i].y); memset(g,0,sizeof(g)); for (int i = 0 ; i < N ; i++) { for(int j = i + 1 ; j < N ; j++) { if (judge(i,j)) g[i][j] = g[j][i] = 1; } } slover.init(); for (int i = 0 ; i < N ; i++) { for (int j = 0 ; j < N ; j++) { if (i != j && g[i][j]) { slover.mat[i][i]++; slover.mat[i][j] = -1; } } } printf("%d\n",slover.det(N - 1) % MOD); } return 0; }
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 20; const int dx[] = {0,0,-1,1}; const int dy[] = {1,-1,0,0}; int N,M; typedef long long type; const long long MOD = 1e9; int id[20][20]; char G[20][20]; int cas; struct Matrix { type mat[MAXN][MAXN]; void init() { memset(mat,0,sizeof(mat)); } type det(int n) { int sign = 0; type ret = 1; for (int i = 0 ; i < n ; i++) { for (int j = i + 1 ; j < n ; j++) { while(mat[j][i]) { type tmp = mat[i][i] / mat[j][i]; for (int k = i ; k < n ; k++) { mat[i][k] = (mat[i][k] - tmp * mat[j][k]); swap(mat[i][k],mat[j][k]); } sign++; } } if (mat[i][i] == 0) return 0; ret = (ret * mat[i][i]); } if (sign & 1) ret = -ret; return ret; } }slover; int g[MAXN][MAXN]; int main() { int T; scanf("%d",&T); while (T--) { int N,M; scanf("%d%d",&N,&M); memset(g,0,sizeof(g)); while (M--) { int u,v; scanf("%d%d",&u,&v); u--; v--; g[u][v] = g[v][u] = 1; } slover.init(); for (int i = 0 ; i < N ; i++) { for (int j = 0 ; j < N ; j++) { if (i != j && g[i][j]) { slover.mat[i][j] = -1; slover.mat[i][i]++; } } } printf("%lld\n",slover.det(N - 1)); } return 0; }
HDU 4408 Minimum Spanning Tree
最小生成树计数。暂时没看太明白
贤贴在这里
生成树计数可以使用Matrix-Tree定理解决，本题最主要的区别是有了一个最小生成树的额外条件。
首先考虑一下如何得到最小生成树。
Kruskal算法的基本思想是，按照边长排序，然后不断将短边加入集合，最终一步如果能成功把n-1条边都加入同一个集合，则找到了最小生成树。在维护集合时，可以使用并查集来快速处理。
如果把Kruskal的过程按照边长划分成多个阶段，实际上是处理了所有短边的连通性之后继续处理下一个长度的边的连通性，并依次继续处理剩下边的连通性。然后我们可以发现，不同长度的边之间的连通性互不影响！！！
假设存在n1条长度为c1的边，n2条长度为c2的边...则Kruskal首先处理c1边的连通性，然后处理c2边的连通性，对于c1边的连通性的处理可能有多种方案，即从n1条边中取出一定数量的边构成最大连通图，但是最终处理完之后的结果对于c2来说是完全一样的。因此算法就出来了，在Kruskal的基础上，使用Matrix-Tree定理处理每个阶段生成树的种数，最后将所有阶段的结果相乘即可。
/* *算法引入： *给定一个含有N个结点M条边的无向图,求它最小生成树的个数t(G); * *算法思想： *抛开“最小”的限制不看,如果只要求求出所有生成树的个数,是可以利用Matrix-Tree定理解决的; *Matrix-Tree定理此定理利用图的Kirchhoff矩阵,可以在O(N3)时间内求出生成树的个数; * *kruskal算法： *将图G={V,E}中的所有边按照长度由小到大进行排序,等长的边可以按照任意顺序; *初始化图G’为{V,Ø},从前向后扫描排序后的边,如果扫描到的边e在G’中连接了两个相异的连通块,则将它插入G’中; *最后得到的图G’就是图G的最小生成树; * *由于kruskal按照任意顺序对等长的边进行排序,则应该将所有长度为L0的边的处理当作一个阶段来整体看待; *令kruskal处理完这一个阶段后得到的图为G0,如果按照不同的顺序对等长的边进行排序,得到的G0也是不同; *虽然G0可以随排序方式的不同而不同,但它们的连通性都是一样的,都和F0的连通性相同(F0表示插入所有长度为L0的边后形成的图); * *在kruskal算法中的任意时刻,并不需要关注G’的具体形态,而只要关注各个点的连通性如何(一般是用并查集表示); *所以只要在扫描进行完第一阶段后点的连通性和F0相同,且是通过最小代价到达这一状态的,接下去都能找到最小生成树; * *经过上面的分析,可以看出第一个阶段和后面的工作是完全独立的; *第一阶段需要完成的任务是使G0的连通性和F0一样,且只能使用最小的代价; *计算出这一阶段的方案数,再乘上完成后续事情的方案数,就是最终答案; * *由于在第一个阶段中,选出的边数是一定的,所有边的长又都为L0; *所以无论第一个阶段如何进行代价都是一样的,那么只需要计算方案数就行了; *此时Matrix-Tree定理就可以派上用场了,只需对F0中的每一个连通块求生成树个数再相乘即可; * *Matrix-Tree定理: *G的所有不同的生成树的个数等于其Kirchhoff矩阵C[G]任何一个n-1阶主子式的行列式的绝对值； *n-1阶主子式就是对于r(1≤r≤n),将C[G]的第r行,第r列同时去掉后得到的新矩阵,用Cr[G]表示; * *算法举例： *HDU4408(Minimum Spanning Tree) * *题目地址： *http://acm.hdu.edu.cn/showproblem.php?pid=4408 * *题目大意： *给定一个含有N个结点M条边的无向图,求它最小生成树的个数,所得结果对p取模; **/ #include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int MAXN = 110; const int MAXM = 1010; struct Edges { int a,b,c; bool operator<(const Edges & x)const { return c<x.c; } }edge[MAXM]; int n,m; int mod; LL f[MAXN],U[MAXN],vist[MAXN];//f,U都是并查集，U是每组边临时使用 LL G[MAXN][MAXN],C[MAXN][MAXN];//G顶点之间的关系，C为生成树计数用的Kirchhoff矩阵 vector<int>V[MAXN];//记录每个连通分量 int Find(int x,LL f[]) { if(x==f[x]) return x; else return Find(f[x],f); } LL det(LL a[][MAXN],int n)//生成树计数:Matrix-Tree定理 { for(int i = 0; i < n ; i++) for(int j = 0; j < n ; j++) a[i][j] %= mod; int ret = 1; for(int i = 1 ; i < n ; i++) { for(int j = i + 1 ; j < n ; j++) while(a[j][i]) { int t = a[i][i] / a[j][i]; for(int k = i ; k < n ; k++) a[i][k] = (a[i][k] - a[j][k] * t) % mod; for(int k = i ; k < n ; k++) swap(a[i][k],a[j][k]); ret = -ret; } if(a[i][i] == 0) return 0; ret = ret * a[i][i] % mod; } return (ret + mod) % mod; } void Solve() { sort(edge,edge + m);//按权值排序 for(int i = 1 ; i <= n ; i++)//初始化并查集 { f[i] = i; vist[i] = 0; } LL Edge = -1;//记录相同的权值的边 LL ans = 1; for(int k = 0 ; k <= m ; k++) { if(edge[k].c != Edge || k == m)//一组相等的边,即权值都为Edge的边加完 { for(int i = 1 ; i <= n ; i++) { if(vist[i]) { LL u = Find(i,U); V[u].push_back(i); vist[i] = 0; } } for(int i = 1 ; i <= n ; i++) //枚举每个连通分量 { if(V[i].size() > 1) { for(int a = 1 ; a <= n ; a++) for(int b = 1 ; b <= n ; b++) C[a][b] = 0; int len = V[i].size(); for(int a = 0 ; a < len ; a++) //构建Kirchhoff矩阵C for(int b = a + 1 ; b < len ; b++) { int a1 = V[i][a]; int b1 = V[i][b]; C[a][b] = (C[b][a] -= G[a1][b1]); C[a][a] += G[a1][b1];//连通分量的度 C[b][b] += G[a1][b1]; } LL ret = (LL)det(C,len); ans=(ans * ret) % mod;//对V中的每一个连通块求生成树个数再相乘 for(int a = 0 ; a < len ; a++) f[V[i][a]] = i; } } for(int i = 1 ; i <= n ; i++) { U[i] = f[i] = Find(i,f); V[i].clear(); } if(k == m) break; Edge = edge[k].c; } int a = edge[k].a; int b = edge[k].b; int a1 = Find(a,f); int b1 = Find(b,f); if(a1 == b1) continue; vist[a1] = vist[b1] = 1; U[Find(a1,U)] = Find(b1,U);//并查集操作 G[a1][b1]++; G[b1][a1]++; } int flag = 0; for(int i = 2; i <= n&& !flag ; i++) if(U[i] != U[i-1]) flag = 1; if(m == 0) flag = 1; printf("%I64d\n",flag ? 0 : ans % mod); } int main() { while(scanf("%d%d%d",&n,&m,&mod) != EOF) { if (n == 0 && m == 0 && mod == 0) break; memset(G,0,sizeof(G)); for(int i = 1 ; i <= n ; i++) V[i].clear(); for(int i = 0 ; i < m ; i++) scanf("%d%d%d",&edge[i].a,&edge[i].b,&edge[i].c); Solve(); } return 0; }
kuangbin带你飞 生成树专题 ： 次小生成树; 最小树形图;生成树计数
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原文地址：http://www.cnblogs.com/Commence/p/4915582.html
