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

題意：在城市A的男孩想去跟城市B的女孩約合，不過他去城市B必須走A->B最短路，並且走過的路不可以再走，問他最多能看這個女孩多少次。 解法：找到A -> B的最短路徑的邊，將其他邊去掉，將最短路上的邊的最大流量設為1，跑一遍最大流，可以得到A -> B的最大流量。。。。 先從A->B做一次spfa，然後再反向建圖從 B->A做一次spfa。然後遍歷原圖每一條邊e(u,v)，若滿足 dis1[u] + dis2[v] + w(u,v) == dis1[B]。 則說明邊 e(u,v) 屬於A - > B 的最短路徑。 （dis1[u] 從A->u的最短路） （dis2[v] 從B->v的最短路） 這道題用dinic的話應該用個好一點的板子，我用的那個模板超時了。。。。 也可以用sap，sap比dinic快好多

spfa + dinic

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
#define pb push_back
#define mp make_pair
#define fi first
#define se second
const int MAXN = 1005;
const int N = 2005;
const int M = N * N;

struct Edge{
    int u,v, w, nxt;
};

int s,t;
int n, m, cnt1,cnt2;
Edge ss[MAXN * MAXN],se[MAXN * MAXN];
int dis1[MAXN],dis2[MAXN],head1[MAXN],head2[MAXN];
bool vis[MAXN];

void init(){
    cnt1 = cnt2 = 0;
    memset(head1,-1,sizeof(head1));
    memset(head2,-1,sizeof(head2));
}

void addEdge(Edge e[],int hh[],int &tt,int u, int v, int w){
    e[tt].u = u;
    e[tt].v = v;
    e[tt].w = w;
    e[tt].nxt = hh[u];
    hh[u] = tt++;
}

struct Dinic{
    int ec, head[N], first[N], que[N], lev[N];
    int Next[M], to[M], v[M];

    void init() {
        ec = 0;
        memset(first, -1, sizeof(first));
    }

    void addEdge(int a,int b,int c) {
        to[ec] = b;
        v[ec] = c;
        Next[ec] = first[a];
        first[a] = ec++;

        to[ec] = a;
        v[ec] = 0;
        Next[ec] = first[b];
        first[b] = ec++;
    }

    int BFS() {
        int kid, now, f = 0, r = 1, i;
        memset(lev, 0, sizeof(lev));
        que[0] = s, lev[s] = 1;
        while (f < r) {
            now = que[f++];
            for (i = first[now]; i != -1; i = Next[i]) {
                kid = to[i];
                if (!lev[kid] && v[i]) {
                    lev[kid] = lev[now] + 1;
                    if (kid == t) return 1;
                    que[r++] = kid;
                }
            }
        }
        return 0;
    }

    int DFS(int now, int sum) {
        int kid, flow, rt = 0;
        if (now == t) return sum;
        for (int i = head[now]; i != -1 && rt < sum; i = Next[i]) {
            head[now] = i;
            kid = to[i];
            if (lev[kid] == lev[now] + 1 && v[i]) {
                flow = DFS(kid, min(sum - rt, v[i]));
                if (flow) {
                    v[i] -= flow;
                    v[i^1] += flow;
                    rt += flow;
                } else lev[kid] = -1;
            }
        }
        return rt;
    }

    int dinic() {
        int ans = 0;
        while (BFS()) {
            for (int i = 0; i <= n; i++) {
                head[i] = first[i];
            }
            ans += DFS(s, inf);
        }
        return ans;
    }
}din;

void spfa(int s,int dd[],int hh[],Edge e[])
{
    memset(vis,false,sizeof(vis));
    for(int i = 1;i <= n;++i){
        dd[i] = inf;
    }
    queue<int>que;
    vis[s] = true;
    dd[s] = 0;
    que.push(s);
    while(!que.empty())
    {
        int tmp = que.front();
        que.pop();
        vis[tmp] = false;
        //cout << hh[tmp] << endl;
        for(int i = hh[tmp];i != -1;i = e[i].nxt)
        {
            int to = e[i].v;
            int val = e[i].w;
            //cout << to << " " << val << endl;
            if(dd[tmp] + val < dd[to]){
                dd[to] = dd[tmp] + val;
                if(!vis[to]){
                    vis[to] = true;
                    que.push(to);
                }
            }
        }
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&m);
        init();
        for(int i = 0;i < m;++i){
            int u,v,w;
            scanf("%d %d %d",&u,&v,&w);
            if(u == v) continue;
            addEdge(ss,head1,cnt1,u,v,w);
            addEdge(se,head2,cnt2,v,u,w);
        }
        scanf("%d %d",&s,&t);
        spfa(s,dis1,head1,ss);
        spfa(t,dis2,head2,se);
//        for(int i = 1;i <= n;++i){
//            printf("%d ",dis1[i]);
//        }
//        printf("\n");
//        for(int i = 1;i <= n;++i){
//            printf("%d ",dis2[i]);
//        }
//        printf("\n");
        din.init();
        for(int i = 0;i < cnt1;++i){
            if(dis1[ss[i].u] + dis2[ss[i].v] + ss[i].w == dis1[t]){
                din.addEdge(ss[i].u,ss[i].v,1);
                //addEdge(edge,head,ecnt,ss[i].v,ss[i].u,0);
                //cout << node[i].u << " " << node[i].v << endl;
            }
        }
        printf("%d\n",din.dinic());
    }
}