How far away ？

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

23 21 2 103 1 151 22 32 21 2 1001 22 1 

 

Sample Output

1025100100 

 

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

#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;  }  

lca模板：

#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h>#include <string.h>#include <limits.h>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;const int maxn=50010;int head[maxn],tol,dis[maxn],fa[maxn][20],dep[maxn];struct node{  int next,to,val;}edge[5*maxn];void addedge(int u,int v,int w){  edge[tol].to=v;  edge[tol].next=head[u];  edge[tol].val=w;  head[u]=tol++;}void bfs(int root){  queue<int> q;  fa[root][0]=root;dep[root]=0;dis[root]=0;  q.push(root);  while(!q.empty()){    int u=q.front();q.pop();    for(int i=1;i<20;i++)fa[u][i]=fa[fa[u][i-1]][i-1];    for(int i=head[u];i!=-1;i=edge[i].next){      int v=edge[i].to;if(v==fa[u][0])continue;      dep[v]=dep[u]+1;dis[v]=dis[u]+edge[i].val;fa[v][0]=u;      q.push(v);    }  }}int lca(int x,int y){  if(dep[x]<dep[y])swap(x,y);  for(int i=0;i<20;i++)if((dep[x]-dep[y])&(1<<i))x=fa[x][i];  if(x==y)return x;  for(int i=0;i<20;i++)if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];  return fa[x][0];}int main(){  //freopen("data.in","r",stdin);  //freopen("data.out","w",stdout);  int T,n,m;   scanf("%d",&T);   while(T--){     scanf("%d%d",&n,&m);     memset(head,-1,sizeof(head));tol=0;     for(int i=1;i<n;i++){       int u,v,w;       scanf("%d%d%d",&u,&v,&w);       addedge(u,v,w);       addedge(v,u,w);     }     bfs(1);     while(m--){       int u,v;       scanf("%d%d",&u,&v);       printf("%d\n",dis[u]+dis[v]-2*dis[lca(u,v)]);     }   }  return 0;}