HDU 3416 Marriage Match IV（最短路，网络流）

题面

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

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

题解

题目大意：
有一个人，特别爱撩妹，现在他在A城市，妹子们在B城市，每次他会从A城市沿着最短的路径到达B城市，并且和一个妹子约会，他每条路只能够走一次，问他最多能够和几个妹子约会？
题解：
首先要确定所有的最短路径上的路，直接用SPFA即可解决（怎么弄自己想）
然后重新连接最短路上的路径，流量为1，求出源点到汇点的最大流即可

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顺便膜拜一下YL大佬，他很强劲的打了我的脸，我现在脸很痛（Cnblogs上不会有这句话的）

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#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;#define MAX 2000#define MAXL 300100#define INF 1000000000inline int read(){       int x=0,t=1;char ch=getchar();       while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();       if(ch=='-'){t=-1;ch=getchar();}       while(ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}       return x*t;}struct Line{       int v,next,w;}e[MAXL],E[MAXL];struct edge{       int v,next,w,fb;}ee[MAXL];int hh[MAX],cntt;int S,T,N,M;int h[MAX],cnt;int H[MAX];inline void Add(int u,int v,int w){       e[cnt]=(Line){v,h[u],w};       E[cnt]=(Line){u,H[v],w};       h[u]=H[v]=cnt++;}int dis1[MAX],dis2[MAX];bool vis[MAX];int level[MAX];void SPFA1(){       for(int i=1;i<=N;++i)vis[i]=false;       for(int i=1;i<=N;++i)dis1[i]=INF;       dis1[S]=0;       queue<int> Q;while(!Q.empty())Q.pop();       Q.push(S);       while(!Q.empty())       {              int u=Q.front();Q.pop();              vis[u]=false;              for(int i=h[u];i;i=e[i].next)              {                    int v=e[i].v,w=e[i].w;                    if(dis1[v]>dis1[u]+w)                    {                            dis1[v]=dis1[u]+w;                            if(!vis[v])                            {                                  vis[v]=true;                                  Q.push(v);                            }                    }              }       }}void SPFA2(){       for(int i=1;i<=N;++i)vis[i]=false;       for(int i=1;i<=N;++i)dis2[i]=INF;       dis2[T]=0;       queue<int> Q;while(!Q.empty())Q.pop();       Q.push(T);       while(!Q.empty())       {              int u=Q.front();Q.pop();              vis[u]=false;              for(int i=H[u];i;i=E[i].next)              {                    int v=E[i].v,w=E[i].w;                    if(dis2[v]>dis2[u]+w)                    {                            dis2[v]=dis2[u]+w;                            if(!vis[v])                            {                                  vis[v]=true;                                  Q.push(v);                            }                    }              }       }}inline void ReAdd(int u,int v,int w){       ee[cntt]=(edge){v,hh[u],w,cnt+1};       hh[u]=cntt++;       ee[cntt]=(edge){u,hh[v],0,cnt-1};       hh[v]=cntt++;}inline void ReBuild(){       for(int i=1;i<=N;++i)       {             for(int j=h[i];j;j=e[j].next)             {                     if(dis1[i]+e[j].w+dis2[e[j].v]==dis1[T])                       ReAdd(i,e[j].v,1);             }       }}inline bool BFS(){       for(int i=1;i<=N;++i)level[i]=0;       level[S]=1;       queue<int> Q;while(!Q.empty())Q.pop();       Q.push(S);       while(!Q.empty())       {             int u=Q.front();Q.pop();             for(int i=hh[u];i;i=ee[i].next)             {                     int v=ee[i].v;                     if(ee[i].w&&!level[v])                     {                             level[v]=level[u]+1;                             Q.push(v);                     }             }       }       return level[T];}int DFS(int u,int f){       if(u==T||f==0)return f;       int re=0;       for(int i=hh[u];i;i=ee[i].next)       {               int v=ee[i].v;               if(ee[i].w&&level[v]==level[u]+1)               {                      int d=DFS(v,min(f,ee[i].w));                      f-=d;re+=d;                      ee[i].w-=d;ee[ee[i].fb].w+=d;                      }       }       return re;}inline int Dinic(){       int re=0;       while(BFS())          re+=DFS(S,INF);       return re;}int main(){       int TT=read();       while(TT--)       {               cnt=cntt=1;               N=read();M=read();               for(int i=1;i<=N;++i)h[i]=H[i]=hh[i]=0;               for(int i=1;i<=M;++i)               {                      int a=read(),b=read(),c=read();                      if(a!=b)                         Add(a,b,c);               }               S=read();T=read();               SPFA1();               SPFA2();               ReBuild();               printf("%d\n",Dinic());       }       return 0;}