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最新推荐文章于 2022-12-04 12:21:01 发布

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本文介绍了一个村庄中关于房屋间距离查询的问题解决方案。通过构建图结构并使用深度优先搜索(DFS)算法来寻找任意两座房屋之间的最短路径，有效地解决了村民提出的各种距离查询需求。

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How far away ？

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11351 Accepted Submission(s): 4139

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

Input

First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

Sample Input

Sample Output

网上代码大多用的c++中的vector，不过貌似不是很好写，第二次手写依旧觉得很神奇

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#include<vector>
using namespace std;
int head[40001],fl,vis[40001],t;
struct node {
    int y,l,next;
}k[80001];
void dfs(int x,int y,int t){
    if(fl)return ;//如果找到路径返回
    if(x==y){
        fl=1;//找到路径
        printf("%d\n",t);
        return ;
    }
    vis[x]=1;
    for(int i=head[x];i!=-1;i=k[i].next){//head数组代表头，依次遍历头所连接的点直到-1也就是访问完
        if(!vis[k[i].y])dfs(k[i].y,y,t+k[i].l);//未访问过的
    }
}
int main(){
    cin>>t;
    while(t--){
        int n,m,h=0,x,y,z;
        scanf("%d%d",&n,&m);
        memset(head,-1,sizeof(head));
        for(int i=0;i<n-1;i++){
            scanf("%d%d%d",&x,&y,&z);
            k[h].y=y;
            k[h].l=z;
            k[h].next=head[x];
            head[x]=h++;
            k[h].y=x;
            k[h].l=z;
            k[h].next=head[y];
            head[y]=h++;
        }
        for(int i=0;i<m;i++){
            scanf("%d%d",&x,&y);
            memset(vis,0,sizeof(vis));
            fl=0;
            dfs(x,y,0);
        }
    }
    return 0;
}