HDU3416(KB11-O spfa+最大流)

Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4381    Accepted Submission(s): 1310

Problem Description

Do not sincere non-interference。
Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can get to city B and make a data with a girl he likes. But there are two problems with it, one is starvae must get to B within least time, it's said that he must take a shortest path. Other is no road can be taken more than once. While the city starvae passed away can been taken more than once.

So, under a good RP, starvae may have many chances to get to city B. But he don't know how many chances at most he can make a data with the girl he likes . Could you help starvae?

 

Input

The first line is an integer T indicating the case number.(1<=T<=65)
For each case,there are two integer n and m in the first line ( 2<=n<=1000, 0<=m<=100000 ) ,n is the number of the city and m is the number of the roads.

Then follows m line ,each line have three integers a,b,c,(1<=a,b<=n,0<c<=1000)it means there is a road from a to b and it's distance is c, while there may have no road from b to a. There may have a road from a to a,but you can ignore it. If there are two roads from a to b, they are different.

At last is a line with two integer A and B(1<=A,B<=N,A!=B), means the number of city A and city B.
There may be some blank line between each case.

 

Output

Output a line with a integer, means the chances starvae can get at most.

 

Sample Input

3
7 8
1 2 1
1 3 1
2 4 1
3 4 1
4 5 1
4 6 1
5 7 1
6 7 1
1 7

6 7
1 2 1
2 3 1
1 3 3
3 4 1
3 5 1
4 6 1
5 6 1
1 6

2 2
1 2 1
1 2 2
1 2

 

Sample Output

2
1
1

 

Author

starvae@HDU

 

Source

HDOJ Monthly Contest – 2010.06.05

 

只有最短路上的边才加入网络，容量都为1，跑最大流

 //2017-08-25
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <queue>

 using namespace std;

 const int N = ;
 const int M = ;
 const int INF = 0x3f3f3f3f;
 int head[N], tot;
 struct Edge{
     int next, to, w;
 }edge[M];

 void add_edge(int u, int v, int w){
     edge[tot].w = w;
     edge[tot].to = v;
     edge[tot].next = head[u];
     head[u] = tot++;
 }

 struct Dinic{
     int level[N], S, T;
     void init(){
         tot = ;
         memset(head, -, sizeof(head));
     }
     void set_s_t(int _S, int _T){
         S = _S;
         T = _T;
     }
     bool bfs(){
         queue<int> que;
         memset(level, -, sizeof(level));
         level[S] = ;
         que.push(S);
         while(!que.empty()){
             int u = que.front();
             que.pop();
             for(int i = head[u]; i != -; i = edge[i].next){
                 int v = edge[i].to;
                 int w = edge[i].w;
                 if(level[v] == - && w > ){
                     level[v] = level[u]+;
                     que.push(v);
                 }
             }
         }
         return level[T] != -;
     }
     int dfs(int u, int flow){
         if(u == T)return flow;
         int ans = , fw;
         for(int i = head[u]; i != -; i = edge[i].next){
             int v = edge[i].to, w = edge[i].w;
             if(!w || level[v] != level[u]+)
               continue;
             fw = dfs(v, min(flow-ans, w));
             ans += fw;
             edge[i].w -= fw;
             edge[i^].w += fw;
             if(ans == flow)return ans;
         }
         if(ans == )level[u] = ;
         return ans;
     }
     int maxflow(){
         int flow = ;
         while(bfs())
           flow += dfs(S, INF);
         return flow;
     }
 }dinic;

 //dis[0][u]表示从起点到u的最短距离，dis[1][u]表示从终点到u的最短距离，cnt[u]记录u入队次数，判负环用
 int dis[][N], cnt[N];
 bool vis[N];
 bool spfa(int s, int n, int op){
     memset(vis, false, sizeof(vis));
     memset(dis[op], INF, sizeof(dis));
     memset(cnt, , sizeof(cnt));
     vis[s] = true;
     dis[op][s] = ;
     cnt[s] = ;
     queue<int> q;
     q.push(s);
     while(!q.empty()){
         int u = q.front();
         q.pop();
         vis[u] = false;
         for(int i = head[u]; i != -; i = edge[i].next){
             int v = edge[i].to;
             int w = edge[i].w;
             if(dis[op][v] > dis[op][u] + w){
                 dis[op][v] = dis[op][u] + w;
                 if(!vis[v]){
                     vis[v] = true;
                     q.push(v);
                     if(++cnt[v] > n)return false;
                 }
             }
         }
     }
     return true;
 }

 struct Line{
     int u, v, w;
 }line[M];

 int main()
 {
     std::ios::sync_with_stdio(false);
     //freopen("inputO.txt", "r", stdin);
     int T, n, m, A, B;
     cin>>T;
     while(T--){
         cin>>n>>m;
         int u, v, w;
         for(int i = ; i < m; i++)
           cin>>line[i].u>>line[i].v>>line[i].w;
         cin>>A>>B;
         dinic.init();
         for(int i = ; i < m; i++)
           add_edge(line[i].u, line[i].v, line[i].w);
         spfa(A, n, );//求从起点开始到各个点的最短路
         dinic.init();
         for(int i = ; i < m; i++)
           add_edge(line[i].v, line[i].u, line[i].w);
         spfa(B, n, );//求从终点开始到各个点的最短路
         dinic.init();
         dinic.set_s_t(A, B);
         for(int i = ; i < m; i++){
             u = line[i].u;
             v = line[i].v;
             w = line[i].w;
             //只有最短路上的边才加入网络
             if(dis[][u]+dis[][v]+w == dis[][B]){
                 add_edge(u, v, );
                 add_edge(v, u, );
             }
         }
         cout<<dinic.maxflow()<<endl;
     }
     return ;
 }