/* ID: wuqi9395@126.com PROG: LANG: C++ */ #include<map> #include<set> #include<queue> #include<stack> #include<cmath> #include<cstdio> #include<vector> #include<string> #include<fstream> #include<cstring> #include<ctype.h> #include<iostream> #include<algorithm> using namespace std; #define LINF (1LL<<60) #define INF (1<<30) #define PI acos(-1.0) #define mem(a, b) memset(a, b, sizeof(a)) #define rep(i, a, n) for (int i = a; i < n; i++) #define per(i, a, n) for (int i = n - 1; i >= a; i--) #define eps 1e-6 #define debug puts("===============") #define pb push_back //#define mp make_pair #define all(x) (x).begin(),(x).end() #define fi first #define se second #define SZ(x) ((int)(x).size()) #define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m) typedef long long ll; typedef unsigned long long ULL; const int maxn = 610; const int maxm = 2000000; int land[210][2], F, P, sumst, sumed; ll mp[210][210], mx; int st, ed, n; int is[210][2]; void get() { scanf("%d%d", &F, &P); n = 1; for (int i = 1; i <= F; i++) { scanf("%d%d", land[i], land[i] + 1); sumst += land[i][0]; sumed += land[i][1]; if (land[i][0]) is[i][0] = n++; if (land[i][1]) is[i][1] = n++; for (int j = 1; j <= F; j++) mp[i][j] = LINF; mp[i][i] = 0; } int u, v, c; for (int i = 0; i < P; i++) { scanf("%d%d%d", &u, &v, &c); if (mp[u][v] > c) mp[u][v] = mp[v][u] = c; //图中两点可能有多条边 } } void floyd(int n, ll mp[][210]) { mx = 0; for (int k = 1; k <= n; k++) { for (int i = 1; i <= n; i++) if (mp[i][k] != LINF) { for (int j = 1; j <= n; j++) { mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]); if (mp[i][j] != LINF) mx = max(mx, mp[i][j]); } } } } struct node { int v; // vertex int cap; // capacity int flow; // current flow in this arc int nxt; } e[maxm * 2]; int g[maxn], cnt; void add(int u, int v, int c) { e[++cnt].v = v; e[cnt].cap = c; e[cnt].flow = 0; e[cnt].nxt = g[u]; g[u] = cnt; e[++cnt].v = u; e[cnt].cap = 0; e[cnt].flow = 0; e[cnt].nxt = g[v]; g[v] = cnt; } void init(ll mid) { mem(g, 0); cnt = 1; st = 0; ed = n++; for (int i = 1; i <= F; i++) if (land[i][0]) add(st, is[i][0], land[i][0]); for (int i = 1; i <= F; i++) if (land[i][1]) add(is[i][1], ed, land[i][1]); for (int i = 1; i <= F; i++) { for (int j = i; j <= F; j++) if (mp[i][j] <= mid) { //j从i开始,因为有草地既有牛,也有避雨点 if (land[i][0] && land[j][1]) add(is[i][0], is[j][1], INF); if (land[j][0] && land[i][1]) add(is[j][0], is[i][1], INF); } } } int dist[maxn], numbs[maxn], q[maxn]; void rev_bfs() { int font = 0, rear = 1; for (int i = 0; i <= n; i++) { //n为总点数 dist[i] = maxn; numbs[i] = 0; } q[font] = ed; dist[ed] = 0; numbs[0] = 1; while(font != rear) { int u = q[font++]; for (int i = g[u]; i; i = e[i].nxt) { if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue; dist[e[i].v] = dist[u] + 1; ++numbs[dist[e[i].v]]; q[rear++] = e[i].v; } } } int maxflow() { rev_bfs(); int u, totalflow = 0; int curg[maxn], revpath[maxn]; for(int i = 0; i <= n; ++i) curg[i] = g[i]; u = st; while(dist[st] < n) { if(u == ed) { // find an augmenting path int augflow = INF; for(int i = st; i != ed; i = e[curg[i]].v) augflow = min(augflow, e[curg[i]].cap); for(int i = st; i != ed; i = e[curg[i]].v) { e[curg[i]].cap -= augflow; e[curg[i] ^ 1].cap += augflow; e[curg[i]].flow += augflow; e[curg[i] ^ 1].flow -= augflow; } totalflow += augflow; u = st; } int i; for(i = curg[u]; i; i = e[i].nxt) if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break; if(i) { // find an admissible arc, then Advance curg[u] = i; revpath[e[i].v] = i ^ 1; u = e[i].v; } else { // no admissible arc, then relabel this vertex if(0 == (--numbs[dist[u]])) break; // GAP cut, Important! curg[u] = g[u]; int mindist = n; for(int j = g[u]; j; j = e[j].nxt) if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]); dist[u] = mindist + 1; ++numbs[dist[u]]; if(u != st) u = e[revpath[u]].v; // Backtrack } } return totalflow; } int main () { get(); floyd(F, mp); if (sumst > sumed) { puts("-1"); return 0; } init(mx); if (maxflow() < sumst) { puts("-1"); return 0; } ll l = 0, r = mx, mid; while(l < r) { mid = (l + r) >> 1; init(mid); if (maxflow() >= sumst) r = mid; else l = mid + 1; } printf("%lld\n", r); return 0; }
POJ 2391 Ombrophobic Bovines (二分 + floyd + 网络流)
原文地址:http://blog.csdn.net/sio__five/article/details/38897013