Description

Problem C: A Walk Through the Forest

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 fromA to B to be progress if there exists a route from B to his home that is shorter than any possible route fromA. 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 intersectionsN, 1 < N ≤ 1000, and the number of paths M. The followingM lines each contain a pair of intersections a  b and an integer distance1 ≤ d ≤ 1000000 indicating a path of length d between intersectiona 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

Output for Sample Input

2
4

题意：Jimmy打算每天沿着一条不同的路走，而且，他只能沿着满足如下条件的道路（A，B）：存在一条从B出发回家的路径，比所以从A出发回家的路径都短，你的任务是计算有多少条不同的路径

思路:所以求回家的最短路径，然后就是求满足dist[A]>dist[B]的所有的可能路径，通过dist[A]>dist[B]我们可以创建一个新图，则题目就是求从起点到终点的路径条数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int MAXN = 1010;
const int INF = 0x3f3f3f3f;

int dp[MAXN], dist[MAXN], cost[MAXN][MAXN];
int vis[MAXN];
int n, m, s, e, w;

void Dijkstra() {
	int Min, cnt;
	for (int i = 1; i <= n; i++)
		dist[i] = cost[2][i];
	vis[2] = 1;
	for (int i = 1; i <= n; i++) {
		Min = INF;
		for (int j = 1; j <= n; j++)
			if (!vis[j] && dist[j] < Min) {
				Min = dist[j];
				cnt = j;
			}
		vis[cnt] = 1;
		if (Min == INF)
			break;
		for (int k = 1; k <= n; k++)
			if (!vis[k] && dist[k] > dist[cnt]+cost[k][cnt])
				dist[k] = dist[cnt] + cost[k][cnt];
	}
}

int dfs(int cur) {
	if (dp[cur] > 0)
		return dp[cur];
	int ans = 0;
	for (int i = 1; i <= n; i++) {
		if (dist[cur] > dist[i] && cost[cur][i] != INF)
			ans += dfs(i);
	}
	return dp[cur] = ans;
}

int main() {
	while (scanf("%d", &n) != EOF && n) {
		scanf("%d", &m);
		memset(cost, INF, sizeof(cost));
		memset(dist, INF, sizeof(dist));
		memset(vis, 0, sizeof(vis));
		memset(dp, 0, sizeof(dp));
		for (int i = 0; i <= n; i++)
			cost[i][i] = 0;
		for (int i = 0; i < m; i++) {
			scanf("%d%d%d", &s, &e, &w);
			cost[s][e] = cost[e][s] = w;
		}
		Dijkstra();
		dp[2] = 1;
		int ans = dfs(1);
		printf("%d\n", ans);
	}	
	return 0;
}