A Walk Through the Forest

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 118 Accepted Submission(s): 57

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

 

Source

University of Waterloo Local Contest 2005.09.24

 

Recommend

Eddy

/*
最短路的条数
*/
#include<bits/stdc++.h>
#define N 1010
#define INF 0x3f3f3f3f
using namespace std;
int n,m;
int mapn[N][N];
int vis[N];
int dis[N];
int dp[N];//表示到点i的最短路有多少条
int x,y,val;
/*
求出到各点的最短路
*/
void dijkstra(int s){
    int i,j,minn,pos;
    memset(vis,,sizeof(vis));
    for(i = ; i<=n; i++)
        dis[i] = mapn[s][i];
    dis[s]=;
    vis[s]=;
    for(i=;i<=n;i++){
        minn=INF;
        for(j=;j<=n;j++){
            if(dis[j]<minn&&!vis[j]){
                minn=dis[pos=j];
            }
        }
        /*
        如果不加这句话minn没有找到的话就是INF就会越界的
        */
        if(minn==INF) break;
        vis[pos]=;
        for(j=;j<=n;j++){
            if(!vis[j]&&dis[pos]+mapn[pos][j]<dis[j]){
                dis[j]=dis[pos]+mapn[pos][j];
            }
        }
    }
}
/*
然后记忆化深搜出结果
*/
int dfs(int v){
    int sum=;
    if(dp[v]!=-) return dp[v];//搜到第v个结点的时候直接拿过来用就行了
    if(v==) return ;
    for(int i=;i<=n;i++){
        /*
        状态转移主要在于dis[v]
        */
        if(mapn[v][i]!=INF&&dis[v]>dis[i])
            sum+=dfs(i);
    }
    dp[v]=sum;
    return dp[v];
}
int main(){
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF&&n){
        for(int i=;i<=n;i++){
            dp[i]=-;
            for(int j=;j<=n;j++){
                mapn[i][j]=INF;
            }
        }
        scanf("%d",&m);
        while(m--){
            scanf("%d%d%d",&x,&y,&val);
            mapn[x][y]=mapn[y][x]=val;
        }
        /*将每点到2的最短距离打到dis数组中去*/
        dijkstra();
        printf("%d\n",dfs());
    }
    return ;
}
/*
36 0 37 34 24
2
4 0 3 3 1 1 2
4
*/