A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8850    Accepted Submission(s): 3267

Problem DescriptionJimmy
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.  InputInput
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.  OutputFor
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 Sample Input5 61 3 21 4 23 4 31 5 124 2 345 2 247 81 3 11 4 13 7 17 4 17 5 16 7 15 2 16 2 10 Sample Output24一个人办公室在点1，家在点2，他要从办公室回家，他从点A到点B当且仅当从B到家的距离必任意一点从A到家的都小，求他回家路线的方案数

//最短路
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
const int N=;
ll g[N][N],vis[N],n,m,k,x,y,z;
ll dp[N],dis[N];
void init()
{
    for(int i=;i<=n;i++)
    {
        for(int j=;j<i;j++)
        {
            g[i][j]=g[j][i]=INF;
        }
        g[i][i]=;
    }
}
void dij(int x)
{
    memset(vis,,sizeof(vis));
    for(int i=;i<=n;i++)
    {
        dis[i]=g[x][i];
    }
    vis[x]=;
    int v=x;
    int minn;
    for(int i=;i<=n;i++)
    {
        minn=INF;
        for(int j=;j<=n;j++)
        {
            if(!vis[j] && minn>dis[j])
            {
                v=j;
                minn=dis[j];
            }
        }
        vis[v]=;
        for(int j=;j<=n;j++)
        {
            if(!vis[j]) dis[j]=min(dis[j],dis[v]+g[v][j]);
        }
    }
}
ll dfs(ll now)
{
    if(dp[now]>) return dp[now];
    if(now==) return ;
    for(int i=;i<=n;i++)
    {
        if(g[now][i]!=INF && dis[now]>dis[i])
            dp[now]+=dfs(i);
    }
    return dp[now];
}
int main()
{
    while(scanf("%lld",&n) && n)
    {
        init();
        scanf("%lld",&m);
        for(int i=;i<m;i++)
        {
            scanf("%lld%lld%lld",&x,&y,&z);
            g[x][y]=g[y][x]=min(g[x][y],z);
        }
        dij();
        memset(dp,,sizeof(dp));
        printf("%lld\n",dfs());
    }
    return ;
}