POJ 1459 最大流

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#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;const int maxn = 110;const int INF = 0x3f3f3f3f;int mp[maxn][maxn], path[maxn], flow[maxn], Start, End, n, nc, np, m, u, v, z;queue<int>q;int bfs(){while (!q.empty()) q.pop();memset(path, -1, sizeof(path));path[Start] = 0;flow[Start] = INF;q.push(Start);while (!q.empty()){int t = q.front(); q.pop();if (t == End) break;for (int i = 0; i <= n; i++)if (i != Start && path[i] == -1 && mp[t][i]){flow[i] = flow[t] < mp[t][i] ? flow[t] : mp[t][i];q.push(i); path[i] = t;}}if (path[End] == -1) return -1;else return flow[n];}int Edmond_Karp(){int max_flow = 0, step, now, pre;while ((step = bfs()) != -1){max_flow += step;now = End;while (now != Start){pre = path[now];mp[pre][now] -= step;mp[now][pre] += step;now = pre;}}return max_flow;}int main(int argc, char const *argv[]){while (~scanf("%d%d%d%d", &n, &np, &nc, &m)){memset(mp, 0, sizeof(mp));while (m--){while (getchar() != '(');scanf("%d,%d)%d", &u, &v, &z);mp[++u][++v] = z;}while (np--){while (getchar() != '(');scanf("%d)%d", &u, &z);mp[0][++u] = z;}while (nc--){while (getchar() != '(');scanf("%d)%d", &u, &z);mp[++u][n + 1] = z;}n++;Start = 0, End = n;printf("%d\n", Edmond_Karp());}return 0;}

输入分别为m个点，a个发电站，b个用户，n条边；

给出每个发电站所能提供的最大流量，给出每个用户所需要的最大流量。

加入超级源点S，连接所有发电站，权值为最大流量，加入超级汇点T，连接所有用户，权值为最大流量。

问题就转换成为普通的最大流问题了。

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