How far away ？(DFS)

最新推荐文章于 2020-07-13 23:03:32 发布

ZSGG_ACM

最新推荐文章于 2020-07-13 23:03:32 发布

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分类专栏： BFS&&DFS 文章标签： dfs

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本文探讨了在特定乡村环境中，如何通过唯一简单路径解决从一个房屋到另一个房屋的距离查询问题，包括输入案例解析和输出样例展示。

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How far away ？

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5766 Accepted Submission(s): 2166

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

Source

ECJTU 2009 Spring Contest

AC代码：

#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>

using namespace std;

const int M = 4 * 1e4 + 100;
typedef long long ll;
typedef pair<int,int> P;
vector<P>G[M];
int vis[M],isGO,e;

void dfs(int x, int cost)
{
    if(isGO || vis[x]) return ;
    if(x == e)//数据太大，要改成全局变量
    {
        isGO = 1;
        printf("%d\n",cost);
        return ;
    }
    vis[x] = 1;
    for(int i = 0; i < G[x].size(); i++)
    {
        P a = G[x][i];
        dfs(a.first,cost + a.second);
    }
    vis[x] = 0;
}

void solve()
{
    int x,y,c,n,m;
    scanf("%d %d",&n,&m);
    for(int i = 1; i < n; i++)
    {
        scanf("%d %d %d",&x,&y,&c);
        G[x].push_back(make_pair(y,c));
        G[y].push_back(make_pair(x,c));
    }
    while(m--)
    {
        isGO = 0;
        scanf("%d %d",&x,&e);
        dfs(x,0);
    }
    for(int i = 1; i <= n; i++) G[i].clear();
}

int main()
{
    int T,cnt = 0;
    scanf("%d",&T);
    while(T--)
    {
        solve();
    }
    return 0;
}