How far away ？ - HDU 2586 LCA 或水题

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How far away ？

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5448 Accepted Submission(s): 2065

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

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

Sample Output

1025100100

思路：用LCA或者因为这道题查询数量比较小，可以暴力。另外后来数据改的会爆栈，所以需要加头文件。

AC代码如下：

#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;struct node{ int v,dis;};vector<node> G[41010];int parent[35][41010],depth[41010],vis[41010],dis[41010],n,m,T,t;void dfs(int v,int p,int d){ parent[0][v]=p;  depth[v]=d;  vis[v]=t;  int i,len=G[v].size();  for(i=0;i<len;i++)   if(vis[G[v][i].v]!=t)   { dis[G[v][i].v]=dis[v]+G[v][i].dis;     dfs(G[v][i].v,v,d+1);   }}void init(int V){ int v,k;  dfs(1,-1,0);  for(k=0;k<30;k++)   for(v=1;v<=V;v++)    if(parent[k][v]<0)     parent[k+1][v]=-1;    else     parent[k+1][v]=parent[k][parent[k][v]];}int lca(int u,int v){ int k;  if(depth[u]>depth[v])   swap(u,v);  for(k=0;k<30;k++)   if((depth[v]-depth[u])>>k &1)    v=parent[k][v];  if(u==v)   return u;  for(k=29;k>=0;k--)   if(parent[k][u]!=parent[k][v])   { u=parent[k][u];     v=parent[k][v];   }  return parent[0][u];}int main(){ int u,v,i,j,k,d;  scanf("%d",&T);  node A;  for(t=1;t<=T;t++)  { scanf("%d%d",&n,&m);    for(i=1;i<=n;i++)     G[i].clear();    for(i=1;i<n;i++)    { scanf("%d%d%d",&u,&v,&d);      A.dis=d;      A.v=v;      G[u].push_back(A);      A.v=u;      G[v].push_back(A);    }    init(n);    while(m--)    { scanf("%d%d",&u,&v);      d=lca(u,v);      printf("%d\n",dis[u]+dis[v]-2*dis[d]);    }    //printf("\n");  }}

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