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比较简单的题了。
只需从左上角到右下角找两条路就可以了。
因为每个点只能走一次，所以拆点，限制流量为1。
因为求的是最大值，所以权值取反求最小值。
因为第一个点和最后一个点经过两次，只算一次，最后要减去。
ps：数组还是开大点好。。。。不知道什么时候就SB了。。。
注意汇点可能不是最后一个点（模板的问题。。
注意初始化的范围。。。。
matrix again只是数据大了，算法不用改。。。无聊。。。。
#include <algorithm> #include <iostream> #include <cstring> #include <string> #include <vector> #include <bitset> #include <cstdio> #include <queue> #include <stack> #include <cmath> #include <list> #include <map> #include <set> #define pk(x) printf("%d\n", x) using namespace std; #define PI acos(-1.0) #define EPS 1E-6 #define clr(x,c) memset(x,c,sizeof(x)) //#pragma comment(linker, "/STACK:102400000,102400000") typedef long long ll; const int MAXV = 2000; const int INF = 1<<30; struct Edge { int to, cap, cost, rev; }; vector<Edge> G[MAXV]; int dist[MAXV], prv[MAXV], pre[MAXV], in[MAXV]; queue<int> que; void addedge(int from, int to, int cap, int cost) { G[from].push_back((Edge){to, cap, cost, G[to].size()}); G[to].push_back((Edge){from, 0, -cost, G[from].size()-1}); } int min_cost_max_flow(int s, int t) { //, int f) { int res = 0; int f = 0; while (1) { //f > 0) { for (int i = 1; i <= t; ++i) dist[i] = INF, in[i] = 0; dist[s] = 0; while (!que.empty()) que.pop(); in[s] = 1; que.push(s); while (!que.empty()) { int u = que.front(); que.pop(); in[u] = 0; for (int i = 0; i < G[u].size(); ++i) { Edge &e = G[u][i]; if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) { dist[e.to] = dist[u] + e.cost; prv[e.to] = u; pre[e.to] = i; if (in[e.to] == 0) { in[e.to] = 1; que.push(e.to); } } } } if (dist[t] == INF) break; //return -1; int d = INF; // d = f; for (int v = t; v != s; v = prv[v]) { d = min(d, G[prv[v]][pre[v]].cap); } f += d; res += d * dist[t]; for (int v = t; v != s; v = prv[v]) { Edge &e = G[prv[v]][pre[v]]; e.cap -= d; G[v][e.rev].cap += d; } } return res; //return f; } int n; int getid1(int x, int y) { return x*n+y+1; } int getid2(int x, int y) { return x*n+y+n*n+1; } int a[30][30]; int main() { while (~scanf("%d", &n)) { for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { scanf("%d", &a[i][j]); } } for (int i = 0; i <= 2*n*n; ++i) G[i].clear(); for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { if (i+j == 0) addedge(getid1(i,j), getid2(i,j), 2, -a[i][j]); else if (i+j == n*2 - 2) addedge(getid1(i,j), getid2(i, j), 2, -a[i][j]); else addedge(getid1(i, j), getid2(i, j), 1, -a[i][j]); if (i+1<n) addedge(getid2(i, j), getid1(i+1, j), 1, 0); if (j+1<n) addedge(getid2(i, j), getid1(i, j+1), 1, 0); } } printf("%d\n", -min_cost_max_flow(getid1(0,0), getid2(n-1,n-1)) - a[0][0] - a[n-1][n-1]); } return 0; }
#include <algorithm> #include <iostream> #include <cstring> #include <string> #include <vector> #include <bitset> #include <cstdio> #include <queue> #include <stack> #include <cmath> #include <list> #include <map> #include <set> #define pk(x) printf("%d\n", x) using namespace std; #define PI acos(-1.0) #define EPS 1E-6 #define clr(x,c) memset(x,c,sizeof(x)) //#pragma comment(linker, "/STACK:102400000,102400000") typedef long long ll; const int MAXV = 610*610*2+2; const int INF = 1<<30; struct Edge { int to, cap, cost, rev; }; vector<Edge> G[MAXV]; int dist[MAXV], prv[MAXV], pre[MAXV], in[MAXV]; queue<int> que; void addedge(int from, int to, int cap, int cost) { G[from].push_back((Edge){to, cap, cost, G[to].size()}); G[to].push_back((Edge){from, 0, -cost, G[from].size()-1}); } int min_cost_max_flow(int s, int t) { //, int f) { int res = 0; int f = 0; while (1) { //f > 0) { for (int i = 1; i <= t; ++i) dist[i] = INF, in[i] = 0; dist[s] = 0; while (!que.empty()) que.pop(); in[s] = 1; que.push(s); while (!que.empty()) { int u = que.front(); que.pop(); in[u] = 0; for (int i = 0; i < G[u].size(); ++i) { Edge &e = G[u][i]; if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) { dist[e.to] = dist[u] + e.cost; prv[e.to] = u; pre[e.to] = i; if (in[e.to] == 0) { in[e.to] = 1; que.push(e.to); } } } } if (dist[t] == INF) break; //return -1; int d = INF; // d = f; for (int v = t; v != s; v = prv[v]) { d = min(d, G[prv[v]][pre[v]].cap); } f += d; res += d * dist[t]; for (int v = t; v != s; v = prv[v]) { Edge &e = G[prv[v]][pre[v]]; e.cap -= d; G[v][e.rev].cap += d; } } return res; //return f; } inline int Scan() { char ch = getchar(); int data = 0; while (ch < ‘0‘ || ch > ‘9‘) ch = getchar(); do { data = data*10 + ch-‘0‘; ch = getchar(); } while (ch >= ‘0‘ && ch <= ‘9‘); return data; } int n; int getid1(int x, int y) { return x*n+y+1; } int getid2(int x, int y) { return x*n+y+n*n+1; } int a[600][600]; int main() { while (~scanf("%d", &n)) { int maxn = 2*n*n; for (int i = 0; i <= maxn; ++i) G[i].clear(); for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { a[i][j] = Scan(); if (i+j == 0 || i+j == n*2-2) addedge(getid1(i,j), getid2(i,j), 2, -a[i][j]); else addedge(getid1(i, j), getid2(i, j), 1, -a[i][j]); if (i+1<n) addedge(getid2(i, j), getid1(i+1, j), 1, 0); if (j+1<n) addedge(getid2(i, j), getid1(i, j+1), 1, 0); } } printf("%d\n", -min_cost_max_flow(getid1(0,0), getid2(n-1,n-1)) - a[0][0] - a[n-1][n-1]); } return 0; }
HDU2686-Matrix & HDU3376-Matrix Again（费用流）
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原文地址：http://www.cnblogs.com/wenruo/p/5873155.html
