Dinic 數組實現

int g[N][N]; /*剩餘流*/<br />bool vis[N]; /*標記數組*/<br />int level[N];<br />int n, m; /*頂點數，邊數*/<br />int src, sink; /*scr源 sink匯*/<br />int build(){<br /> /* 建圖 */<br />}<br />int find(int x) {<br /> for (int i = 1; i <= n; i++) {<br /> if (g[x][i] != 0 && level[i] == level[x] + 1)<br /> return i;<br /> }<br /> return 0;<br />}<br />int bfs() {<br /> queue<int> que;<br /> bool flag = 0;<br /> for (int i = 1; i <= n; i++) {<br /> level[i] = inf;<br /> vis[i] = false;<br /> }<br /> que.push(src);<br /> vis[src] = true;<br /> level[src] = 0;<br /> while (!que.empty()) {<br /> int temp = que.front();<br /> que.pop();<br /> for (int i = 1; i <= n; i++) {<br /> if (g[temp][i] != 0 && !vis[i]) {<br /> vis[i] = true;<br /> level[i] = level[temp] + 1;<br /> que.push(i);<br /> }<br /> }<br /> if (temp == sink) flag = true;<br /> }<br /> return flag;<br />}<br />int dinic() {<br /> int u, v, capflow, last;<br /> vector<int> path;<br /> int maxflow = 0;<br /> while (bfs()) {<br /> path.clear();<br /> path.push_back(src);<br /> while (find(src) != 0) {<br /> u = path.back();<br /> if (u != sink) {<br /> if (v = find(u)) path.push_back(v);<br /> else { path.pop_back(); level[u] = inf; }<br /> } else {<br /> capflow = inf;<br /> for (int i=0;i<path.size();i++) {<br /> u = path[i];<br /> if((i+1)==path.size()) break;<br /> v = path[i+1];<br /> if (g[u][v] < capflow)<br /> capflow = g[u][v];<br /> }<br /> last = -1;<br /> maxflow += capflow;<br /> for (int i=0;i<path.size();i++){<br /> u = path[i];<br /> if ((i+1) == path.size()) break;<br /> v = path[i+1];<br /> g[u][v] -= capflow;<br /> g[v][u] += capflow;<br /> if (g[u][v] == 0 && last == -1)<br /> last = u;<br /> }<br /> while (path.back() != last) path.pop_back();<br /> }<br /> }<br /> }<br /> return maxflow;<br />}