Problem Description：

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： 

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.  

Output： 

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 

Sample Input： 

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 

Sample Output： 

2

4 

解题思路： 

这道题很容易误解成求1和2之间的最短路数目，我刚开始就是这样理解的，导致输出结果一直不对，后来仔细读了读题，才发现题目的意思是：如果A到终点的距离大于B到终点的距离，那么我们认为A到B是满足条件的路径，也就是说我们可以先将终点2作为起点，用Dijkstra求出单元最短路，得到每个点到终点的距离，然后记忆化搜索，寻找满足条件的路径数目即可！！！（真的是一道非常nice的题目）

程序代码：  
#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f
int map[1001][1001],book[1001],dis[1001];
int p[1001],q[1001];
int n,m;
void Dijkstra(int v)
{
	int u,min;
	memset(book,0,sizeof(book));
	for(int i=1;i<=n;i++)
		dis[i]=map[v][i];
	book[v]=1;
	dis[v]=0;
	for(int i=1;i<n;i++)
	{
		min=INF;
		for(int j=1;j<=n;j++)
		{
			if(book[j]==0&&dis[j]<min)
			{
				u=j;
				min=dis[j];
			}
		}
		book[u]=1;
		for(int v=1;v<=n;v++)
		{
			if(map[u][v]<INF)
			{
				if(dis[v]>dis[u]+map[u][v])
					dis[v]=dis[u]+map[u][v];
			}
		}
	}
}
int dfs(int x)
{
	if(p[x])//如果走过，返回走过此路的路数 
		return p[x];
	if(x==2)//走到终点2，返回1，代表路数+1 
		return 1;
	int sum=0;
	for(int i=1;i<=n;i++)
	{
		if(dis[x]>dis[i]&&map[x][i]<INF)
			sum+=dfs(i);//加上路径数目 
	}
	return p[x]=sum;//返回总的数目 
}
int main()
{
	int a,b,c;
	while(scanf("%d %d",&n,&m)!=EOF)
	{
		if(n==0)
			break;
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				if(i==j)
					map[i][j]=0;
				else
					map[i][j]=INF;
			}
		}
		for(int i=0;i<m;i++)
		{
			scanf("%d %d %d",&a,&b,&c);
			if(map[a][b]>c)
				map[a][b]=map[b][a]=c;
		}
		Dijkstra(2);
		memset(p,0,sizeof(p));
		printf("%d\n",dfs(1));
	}
	return 0;
}