Marriage Match IV(最短路径+最大流）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3416

题目大意：求最短路的个数

题目思路：先用SPFA求出最短路径，然后用SAP算法求出最大网络流，即最短的个数，以最短路径上的边建图，用邻接链表存边

题目：

Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1478    Accepted Submission(s): 412

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
  

 

#include <iostream>
#include <queue>
#include <vector>
#include<string.h>
#include<stdio.h>
using namespace std;
const int inf=0x3f3f3f3f;
struct rec
{
    int u,v,w,link;
} edge[300005];
struct re
{
    int v,c;
};
vector<re> E[1005];
bool vis[1005];
int head[1005],pre[1005],cur[1005],dis[1005],gap[1005];
int n,m,num,st,ed;
inline void addedge(int u,int v,int w)
{
    edge[num].u=u;
    edge[num].v=v;
    edge[num].w=w;
    edge[num].link=head[u];
    head[u]=num++;
    edge[num].u=v;
    edge[num].v=u;
    edge[num].w=0;
    edge[num].link=head[v];
    head[v]=num++;
}
void SPFA()
{
    int i,v,u;
    memset(vis,0,sizeof(vis));
    for(i=1; i<=n; i++)
        dis[i]=inf;
    queue<int> Q;
    dis[st]=0;
    Q.push(st);
    while(!Q.empty())
    {
        u=Q.front();
        Q.pop();
        vis[u]=0;
        for(i=0; i<E[u].size(); i++)
        {
            v=E[u][i].v;
            if(dis[v]>dis[u]+E[u][i].c)
            {
                dis[v]=dis[u]+E[u][i].c;
                if(!vis[v])
                {
                    vis[v]=1;
                    Q.push(v);
                }
            }
        }
    }
}
int max_flow()
{
    int i,k,u,v,aug=inf,ans=0;
    memset(dis,0,sizeof(dis));
    memset(gap,0,sizeof(gap));
    gap[0]=n;
    u=pre[st]=st;
    for(i=1; i<=n; i++)
        cur[i]=head[i];
    while(dis[st]<n)
    {
loop:
        for(i=cur[u]; i; i=cur[u]=edge[i].link)
        {
            v=edge[i].v;
            if(dis[u]==dis[v]+1 && edge[i].w)
            {
                pre[v]=u;
                u=v;
                if(aug>edge[i].w)
                    aug=edge[i].w;
                if(v==ed)
                {
                    for(u=pre[v]; v!=st; v=u,u=pre[u])
                    {
                        edge[cur[u]].w-=aug;
                        edge[cur[u]^1].w+=aug;
                    }
                    ans+=aug;
                    aug=inf;
                }
                goto loop;
            }
        }
        k=n;
        for(i=head[u]; i; i=edge[i].link)
        {
            v=edge[i].v;
            if(k>dis[v] && edge[i].w)
            {
                cur[u]=i;
                k=dis[v];
            }
        }
        if(--gap[dis[u]]==0)
            break;
        gap[dis[u]=k+1]++;
        u=pre[u];
    }
    return ans;
}
int main()
{
    int T,i,j,u,v,c;
    re tmp;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=1; i<=n; i++)
            E[i].clear();
        num=2;
        memset(head,0,sizeof(head));
        for(i=1; i<=m; i++)
        {
            scanf("%d%d%d",&u,&v,&c);
            if(u==v)
                continue;
            tmp.v=v;
            tmp.c=c;
            E[u].push_back(tmp);
        }
        scanf("%d%d",&st,&ed);
        SPFA();
        for(i=1; i<=n; i++)
            for(j=0; j<E[i].siz e(); j++)
                if(dis[E[i][j].v]==dis[i]+E[i][j].c)
                    addedge(i,E[i][j].v,1);
        printf("%d\n",max_flow());
    }
    return 0;
}