HDU2586How far away ?

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   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

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define eps 1e-8
#define ll long long
const int inf = 0x3f3f3f3f;
const long long mod=1e9+7;
const int N=40000+20;
using namespace std;
struct node
{
    int v,c;
}tmp;
vectorq[N];
vectore[N];
int n,m,t;
int fa[N],sum[N],vis[N],ans[N];
void init()
{
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
        fa[i]=i;
}
int find(int x)
{
    while(x!=fa[x])
        x=fa[x];
    return x;
}
int dfs(int u,int pre,int val)
{
    sum[u]=val;
    int sz=e[u].size();
    for(int i=0;i




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