UVA 10917 Walk Through the Forest

最新推荐文章于 2018-08-12 20:22:23 发布

码代码的猿猿的AC之路

最新推荐文章于 2018-08-12 20:22:23 发布

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本文介绍了一个经典的图论问题——寻找从办公室到家的不同路径数量，通过最短路径算法结合动态规划来解决。输入包括交叉路口数量及路径详情，输出所有可能路径的数量。

最短路+DP

Walk Through the Forest

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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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 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

Output for Sample Input

2
4

(apologies to) Richard Krueger

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=1010,INF=0x3f3f3f3f;

int g[maxn][maxn],n,m,dist[maxn];
bool vis[maxn];

void dijkstra(int st)
{
    memset(vis,0,sizeof(vis));
    memset(dist,63,sizeof(dist));
    dist[st]=0;

    for(int i=1;i<=n;i++)
    {
        int mark=-1,mindist=INF;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dist[j]<mindist)
            {
                mindist=dist[j];
                mark=j;
            }
        }
        if(mark==-1) break;
        vis[mark]=true;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dist[j]>dist[mark]+g[mark][j])
                dist[j]=dist[mark]+g[mark][j];
        }
    }
}

int dp[maxn];

int dfs(int u)
{
    if(dp[u]!=-1)
        return dp[u];
    int res=0;
    for(int i=1;i<=n;i++)
    {
        if(g[u][i]!=INF&&dist[i]<dist[u])
            res+=dfs(i);
    }
    return dp[u]=res;
}

int main()
{
    while(scanf("%d",&n)!=EOF&&n)
    {
        scanf("%d",&m);
        memset(g,63,sizeof(g));
        for(int i=0;i<m;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            g[a][b]=g[b][a]=c;
        }
        dijkstra(2);
        memset(dp,-1,sizeof(dp));
        dp[2]=1;
        printf("%d\n",dfs(1));
    }
    return 0;
}