HDU 1142 A Walk Through the Forest锛堟渶鐭矾+璁板繂鍖栨悳绱級

Posted 2020-12-29

鏍囩锛?a href='http://www.mamicode.com/so/1/memset' title='memset'>memset   contest   esc   pos   pat   home   tin   鍗曟簮鏈€鐭矾   ice   

A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10172    Accepted Submission(s): 3701

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

 

 

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code锛?/span>

#include<iostream>
#include<stdio.h>
#include<memory.h>
using namespace std;
#define max_v 1005
#define INF 9999999
int n,m;
int vis[max_v];
int dis[max_v];
int e[max_v][max_v];
int dp[max_v];
void init()
{
    memset(vis,0,sizeof(vis));
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=n;j++)
        {
            e[i][j]=INF;
        }
        dp[i]=-1;
        dis[i]=INF;
    }
}
void Dijkstra(int s)
{
    for(int i=1;i<=n;i++)
        dis[i]=e[s][i];
    dis[s]=0;
    for(int i=1;i<=n;i++)
    {
        int index,mindis=INF;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dis[j]<mindis)
            {
                mindis=dis[j];
                index=j;
            }
        }
        vis[index]=1;
        for(int j=1;j<=n;j++)
            if(dis[index]+e[index][j]<dis[j])
                dis[j]=dis[index]+e[index][j];
    }
}
int dfs(int v)
{
    if(dp[v]!=-1)
        return dp[v];
    if(v==2)
        return 1;
    int sum=0;
    for(int i=1;i<=n;i++)
        if(dis[v]>dis[i]&&e[v][i]!=INF)
             sum+=dfs(i);
    dp[v]=sum;
    return dp[v];
}
int main()
{
    while(~scanf("%d",&n))
    {
        if(!n)
            break;
        scanf("%d",&m);
        init();
        for(int i=0;i<m;i++)
        {
            int x,y,z;
            scanf("%d %d %d",&x,&y,&z);
            if(e[x][y]>z)
                e[x][y]=e[y][x]=z;
        }
        Dijkstra(2);
        printf("%d
",dfs(1));
    }
    return 0;
}