学步园

Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To make things even nicer, his office is on one side of a forest, and his
house is on the other. A nice walk through the forest, seeing the birds and chipmunks is quite enjoyable.

The forest is beautiful, and Jimmy wants to take a different route everyday. He also wants to get home before dark, so he always takes a path to make progress towards his house. He considers taking a path from A to B to be progress if there exists a route from
B to his home that is shorter than any possible route from A. Calculate how many different routes through the forest Jimmy might take.

Input contains several test cases followed by a line containing 0. Jimmy has numbered each intersection or joining of paths starting with 1. His office is numbered 1, and his house is numbered 2. The first line of each test case gives
the number of intersections N, 1 < N ≤ 1000, and the number of paths M. The following M lines each contain a pair of intersections a b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between intersection a and a different intersection
b. Jimmy may walk a path any direction he chooses. There is at most one path between any pair of intersections.

For each test case, output a single integer indicating the number of different routes through the forest. You may assume that this number does not exceed 2147483647

5 6
1 3 2
1 4 2
3 4 3
1 5 12
4 2 34
5 2 24
7 8
1 3 1
1 4 1
3 7 1
7 4 1
7 5 1
6 7 1
5 2 1
6 2 1
0

2
4
题目意思注意点：题目不是求最短路有多少条，而是，假如有一条A到B的路，而B到家的最短路比A到家的最短路要小，那么就可以从A到B，B到家，这可以算作一条路可走。求共有多少条这样的路从1点到家（2）。
解题：先算出每个点到点2的最短路算出来，然后用深搜+标记，减少时间，算出共有多少条路可走到家。
#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;

typedef struct n1
{
    int  distens,flog;
}node;
node N[1005];
int map[1005][1005],k;
int direct[1005];
void set(int n)
{
    int i,j,m,n1,n2,d;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
        {
            map[i][j]=-1;
        }
         N[i].distens=10000000;N[i].flog=0;direct[i]=0;
    }

    scanf("%d",&m);
    while(m--)
    {
        scanf("%d%d%d",&n1,&n2,&d);
        if(map[n1][n2]!=0||map[n1][n2]>d)
        map[n1][n2]=map[n2][n1]=d;
    }
}
void spfa(int n)
{
    queue<int> Q;
    int now;
    int i;
     N[2].distens=0;N[2].flog=1;
    Q.push(2);
    while(!Q.empty())
    {
        now=Q.front();
        Q.pop();
        N[now].flog=0;
        //if(q.x==1)break;
        for(i=1;i<=n;i++)
        if(map[now][i]!=-1)
        {
            if(N[i].distens>N[now].distens+map[now][i])
            {
                N[i].distens=N[now].distens+map[now][i];
                if(N[i].flog==0)
                {
                    N[i].flog=1;
                    Q.push(i);
                }
            }
        }
    }
}
int DFS(int now,int n)
{
    int i;
    if(direct[now]>0)
    return direct[now];
    if(now==2)
    {
        return 1;
    }
    for(i=1;i<=n;i++)
    if(map[now][i]!=-1&&N[now].distens>N[i].distens)
    {
        direct[now]+=DFS(i,n);
    }

   return direct[now];
}
int main()
{
    int n;
    while(scanf("%d",&n)>0&&n)
    {
        set(n);
        spfa(n);
        k=DFS(1,n);
        printf("%d\n",k);
    }
}