hdu 2586 How far away ？ DFS大水题

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2935    Accepted Submission(s): 1095

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

   
   
   
   

    
    
    
    2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1
   
   
   
   

 

 

Sample Output

   
   
   
   

    
    
    
    10
25
100
100
   
   
   
   

 

 

Source

ECJTU 2009 Spring Contest

 

 

Recommend

lcy

 

 

题目大意：一个村子里有n个房子，这n个房子用n-1条路连接起来，接下了有m次询问，每次询问两个房子a,b之间的距离是多少。

从某点到某点有且只有一条路

 

思路 ： DFS  一直D下去 直到找到我们要到的地方  由于无法保存那么大的map路径  可以用vector

#include<stdio.h>
#include<vector>
#include<string.h>
using namespace std;
struct haha
{
    int pos;
    int val;
}temp,q;
vector<struct haha>a[44444];
int n,m,flag,e,vis[444444];
void DFS(int s,int ans)
{
    //printf("s=%d\n",s);
    int size,i;
    if(vis[s]) return ;
    if(flag) return ;
    if(s==e) {printf("%d\n",ans);flag=1;return;}
     if(a[s].empty())  return ;
     else
     {
    vis[s]=1;
     size=a[s].size();
     for(i=0;i<size;i++)
        DFS(a[s][i].pos,ans+a[s][i].val);
     vis[s]=0;
     }
}
int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        int i,j,x,y,z;
        scanf("%d %d",&n,&m);
        for(i=0;i<n-1;i++)
        {
                scanf("%d %d %d",&x,&y,&z);
                q.pos=y;q.val=z;
                a[x].push_back(q);
                q.pos=x;q.val=z;
                a[y].push_back(q);
        }
        for(j=0;j<m;j++)
        {
            memset(vis,0,sizeof(vis));
            flag=0;
            int s;
            scanf("%d %d",&s,&e);
            DFS(s,0);
        }
        for(i=0;i<n;i++)
        {
            a[i].clear();
        }
    }
      return 0;
}