/*0為源點,n-1為匯點,返回最大流*/<br />int edge[N][N];<br />int chk[N], n,m;<br />int father[N];<br />int Ford_Fulkerson() {<br /> while (1) {<br /> queue<int> q;<br /> memset(chk, 0, sizeof (chk));<br /> memset(father, -1, sizeof (father));<br /> int now;<br /> chk[0] = 1;<br /> q.push(0);<br /> while (!q.empty()) {<br /> now = q.front(); q.pop();<br /> if (now == n - 1) break;<br /> for (int i = 0; i < n; i++)<br /> if (edge[now][i] && !chk[i]) {<br /> father[i] = now;<br /> chk[i] = 1;<br /> q.push(i);<br /> }<br /> }<br /> if (!chk[n - 1]) break;<br /> int u, min = inf;<br /> for (u = n - 1; u; u = father[u])<br /> if (edge[father[u]][u] < min) min = edge[father[u]][u];<br /> for (u = n - 1; u; u = father[u]) {<br /> edge[father[u]][u] -= min;<br /> edge[u][father[u]] += min;<br /> }<br /> }<br /> int ans = 0;<br /> for (int i = 0; i < n; i++)<br /> ans += edge[n - 1][i];<br /> return ans;<br />}