HDU 5889 Barricade（最短路建图跑网络流）

这是一道玄学题目。

跑最短路很简单，跑网络流也是模版而已。

但是我跑完最短路后从n开始向前bfs找满足dis[u] == dis[v] + 1的边，然后加入网络流的图里面。这样做会TLE。

然后想了好久实在不知道搞什么鬼，看了看别人代码，别人是直接把所有边for一遍，只要满足dis[u] + 1 == dis[v]的，都加入网络流里面。这样当然也可以，我改了改，15ms就过了。但是这bfs也不至于超时吧？？？好浪费时间啊，这啥破题啊，本来一小时写完了搞了七个小时，mmp

代码如下：

#include<iostream>#include<cstdio>#include<vector>#include<queue>#include<utility>#include<stack>#include<algorithm>#include<cstring>#include<string>#include<cmath>using namespace std;const int maxn = 2005;const int maxm = 50005;const int INF = 0x3f3f3f3f;typedef pair<int, int> pii;int n, m;int head[maxn],cur[maxn],nx[maxm<<1],to[maxm<<1],flow[maxm<<1], ppp=0;int dis[maxn];struct Dinic{int s, t;int ans;void init() {memset(head, -1, sizeof(head));ppp = 0;}void AddEdge(int u, int v, int c){to[ppp]=v;flow[ppp]=c;nx[ppp]=head[u];head[u]=ppp++;swap(u,v);to[ppp]=v;flow[ppp]=0;nx[ppp]=head[u];head[u]=ppp++;}bool BFS(){memset(dis, -1, sizeof(dis));dis[s] = 1; queue<int> Q;Q.push(s);while(!Q.empty()){int x = Q.front();Q.pop();for(int i = head[x]; ~i; i = nx[i]){if(flow[i] && dis[to[i]] == -1){dis[to[i]] = dis[x] + 1;Q.push(to[i]);}}}return dis[t] != -1;}int DFS(int x, int maxflow){if(x == t || !maxflow){ans += maxflow;return maxflow;}int ret = 0, f;for(int &i = cur[x]; ~i; i = nx[i]){if(dis[to[i]] == dis[x] + 1 && (f = DFS(to[i], min(maxflow, flow[i])))){ret += f;flow[i] -= f;flow[i^1] += f;maxflow -= f;if(!maxflow)break;}}return ret;}int solve(int source, int tank){s = source;t = tank;ans = 0;while(BFS()){memcpy(cur, head, sizeof(cur));DFS(s, INF);}return ans;}}dinic;vector<pii>edge[maxn];bool flag[maxn];void spfa() {priority_queue <pii, vector<pii>, greater<pii> > q;int u, v;pii tmp;memset(dis, 0x3f, sizeof(dis));dis[1] = 0;q.push(pii(0, 1));while(!q.empty()) {tmp = q.top();u = tmp.second;q.pop();if(dis[u] < tmp.first)continue;if(u == n)break;for(int i = 0; i < edge[u].size(); i++) {tmp = edge[u][i];v = tmp.first;if(dis[v] > dis[u] + 1) {dis[v] = dis[u] + 1;q.push(pii(dis[v], v));}}}}//void findpath() {//int u, v;//queue <int> q;//q.push(n);//while(!q.empty()) {//u = q.front();//q.pop();//if(u == 1)//break; ////for(int i = 0; i < edge[u].size(); i++) {//v = edge[u][i].first;//if(dis[v] + 1 == dis[u]) {//dinic.AddEdge(v, u, edge[u][i].second);//q.push(v);//}//}//}//}void findpath() {int v;for(int i = 1; i <= n; i++) {for(int j = 0; j < edge[i].size(); j++) {v = edge[i][j].first;if(dis[v] == dis[i] + 1) {dinic.AddEdge(i, v, edge[i][j].second);}}}}int main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);//    freopen("out.txt", "w", stdout);#endifint T;scanf("%d", &T);while(T--) {scanf("%d%d", &n, &m);for(int i = 0; i < maxn; i++) {edge[i].clear();}dinic.init();int u, v, val;while(m--) {scanf("%d%d%d", &u, &v, &val);edge[u].push_back(pii(v, val));edge[v].push_back(pii(u, val));}spfa();findpath();int ans = dinic.solve(1, n);printf("%d\n", ans);}return 0;}