hdu 1142 A Walk Through the Forest

最新推荐文章于 2021-02-18 17:24:09 发布

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#最短路 #DFS

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本文介绍了一个关于寻找森林中从办公室到家的最短路径数量的问题。通过Dijkstra算法找到最短路径后，再使用深度优先搜索（DFS）来计算可能的不同路径的数量。

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A Walk Through the Forest

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9269 Accepted Submission(s): 3419

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

Sample Output

2
4

题意：从1走到2有最短路为定值，求最短路有几条

思路：先求出最短路，然后搜索求出最短路的条数

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=1005;
const int inf=0x3f3f3f3f;
int cost[N][N];
int dp[N],path[N];
int n,m;
void Dijkstra(int s)
{
    int vis[N]={0};
    int i,j,pos,mi;
    for(i=0;i<=n;i++)
        dp[i]=cost[s][i];
    dp[s]=0;
    vis[s]=1;
    for(i=1;i<=n;i++)
    {
        mi=inf;
        pos=0;//一定要加这个*********
        for(j=1;j<=n;j++)
        {
            if(dp[j]<mi&&!vis[j])
            {
                mi=dp[j];
                pos=j;
            }
        }
        vis[pos]=1;
        for(j=1;j<=n;j++)
        {
            if(dp[j]>dp[pos]+cost[pos][j]&&!vis[j])
                dp[j]=dp[pos]+cost[pos][j];
        }
    }
}
int DFS(int s)
{
    if(path[s]!=-1)
        return path[s];
    if(s==2)
        return 1;
    path[s]=0;
    for(int i=1;i<=n;i++)
    {
        if(dp[i]<dp[s]&&cost[i][s]!=inf)
            path[s]+=DFS(i);
    }
    return path[s];
}
int main()
{
    while(scanf("%d",&n),n)
    {
        scanf("%d",&m);
        //初始化
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
                cost[i][j]=cost[j][i]=dp[i]=inf;
            cost[i][i]=0;
        }
        int x,y,z;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            cost[x][y]=cost[y][x]=z;
        }
        Dijkstra(2);
        memset(path,-1,sizeof(path));
        printf("%d\n",DFS(1));
        //printf("#%d\n",dp[1]);
    }
    return 0;
}

参考博客

http://blog.youkuaiyun.com/libin56842/article/details/17270239