Sightseeing

Tour operator Your Personal Holiday organises guided bus trips across the Benelux. Every day the bus moves from one city S to another city F. On this way, the tourists in the bus can see the sights alongside the route travelled. Moreover, the bus makes a number of stops (zero or more) at some beautiful cities, where the tourists get out to see the local sights.

Different groups of tourists may have different preferences for the sights they want to see, and thus for the route to be taken from S to F. Therefore, Your Personal Holiday wants to offer its clients a choice from many different routes. As hotels have been booked in advance, the starting city S and the final city F, though, are fixed. Two routes from S to F are considered different if there is at least one road from a city A to a city B which is part of one route, but not of the other route.

There is a restriction on the routes that the tourists may choose from. To leave enough time for the sightseeing at the stops (and to avoid using too much fuel), the bus has to take a short route from S to F. It has to be either a route with minimal distance, or a route which is one distance unit longer than the minimal distance. Indeed, by allowing routes that are one distance unit longer, the tourists may have more choice than by restricting them to exactly the minimal routes. This enhances the impression of a personal holiday.



For example, for the above road map, there are two minimal routes from S = 1 to F = 5: 1 → 2 → 5 and 1 → 3 → 5, both of length 6. There is one route that is one distance unit longer: 1 → 3 → 4 → 5, of length 7.

Now, given a (partial) road map of the Benelux and two cities S and F, tour operator Your Personal Holiday likes to know how many different routes it can offer to its clients, under the above restriction on the route length.

2
5 8
1 2 3
1 3 2
1 4 5
2 3 1
2 5 3
3 4 2
3 5 4
4 5 3
1 5
5 6
2 3 1
3 2 1
3 1 10
4 5 2
5 2 7
5 2 7
4 1

3
2

//62MS	444K
#include<stdio.h>
#include<string.h>
#define M 10007
#define inf 0x3f3f3f
int map[M][M],dist[M][2],count[M][2],head[M];
bool vis[M][2];
int n,m,num;
struct E
{
    int v,w,to;
} edg[M<<4];
void addedge(int u,int v,int w)
{
    edg[num].v=v;
    edg[num].w=w;
    edg[num].to=head[u];
    head[u]=num++;
}
void dijkstra(int s)
{
    int pos,min,k;
    for(int i=1; i<=n; i++)
    {
        dist[i][0]=inf;
        dist[i][1]=inf;
    }
    memset(vis,false,sizeof(vis));
    count[s][0]=1;
    dist[s][0]=0;
    for(int i=1; i<2*n; i++)//每个点要用两次
    {
        pos=-1;
        min=inf;
        for(int j=1; j<=n; j++)
            if(!vis[j][0]&&min>dist[j][0])//找最短路,每个节点先更新最短，然后次短
            {
                k=0;
                min=dist[j][0];
                pos=j;
            }
            else if(!vis[j][1]&&min>dist[j][1])//找次短路
            {
                k=1;
                min=dist[j][1];
                pos=j;
            }
        if(pos==-1)break;//如果找不到最短或者次短就结束
        vis[pos][k]=1;
        for(int j=head[pos]; j!=-1; j=edg[j].to)
        {
            int v=edg[j].v;
            int len=dist[pos][k]+edg[j].w;
            if(len<dist[v][0])//如果新路径比最短路要短
            {
                dist[v][1]=dist[v][0];//更新最短路径的长度
                count[v][1]=count[v][0];//更新最短路径的条数
                dist[v][0]=len;//更新次短路径的长度
                count[v][0]=count[pos][k];//更新次短路径的条数
            }
            else if(len==dist[v][0])//如果新路径等于最短路径的长度
                count[v][0]+=count[pos][k];//更新最短路径的条数
            else if(len<dist[v][1])//如果新路径比最短路长且比次短路短
            {
                dist[v][1]=len;//更新次短路的长度
                count[v][1]=count[pos][k];//更新次短路径的条数
            }
            else if(len==dist[v][1])//如果新路径长度等于次短路径的长度
                count[v][1]+=count[pos][k];//更新次短路径的条数
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        num=0;
        memset(head,-1,sizeof(head));
        scanf("%d%d",&n,&m);
        int a,b,c,s,t;
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&a,&b,&c);//因为题目中可能有重变，所以要用临街表存储
            addedge(a,b,c);
        }
        scanf("%d%d",&s,&t);
        dijkstra(s);
        int ans=count[t][0];
        if(dist[t][1]==dist[t][0]+1)//如果次短路比最短路长度多1
            ans+=count[t][1];//加上次短路的条数
        printf("%d\n",ans);
    }
}