[HDU]P2586 How far away?[LCA]

[HDU]P2586

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18675    Accepted Submission(s): 7274

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

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---

这道题就是很裸的LCA，主要是练一下倍增，今天考试一道有关LCA的，我用树剖打竟然T了？（感觉效率有保证，不知道是不是数据问题）

可恶啊，打了很久诶，于是就来学习一下倍增。

代码：

 //2017.11.7
 //lca
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 inline int read();
 namespace lys{
      ;
     struct edge{
         int to;
         int next;
         int w;
     }e[N*];
     ],dis[N][],dep[N],pre[N];
     int n,m,cnt;
     void swap(int &a,int &b){int t=a;a=b;b=t;}
     void add(int x,int y,int w){
         e[++cnt].to=y;e[cnt].next=pre[x];pre[x]=cnt;e[cnt].w=w;
         e[++cnt].to=x;e[cnt].next=pre[y];pre[y]=cnt;e[cnt].w=w;
     }
     void dfs(int node,int deep){
         dep[node]=deep;
         int i,v;
         ;i<=;i++) anc[node][i]=anc[anc[node][i-]][i-],dis[node][i]=dis[node][i-]+dis[anc[node][i-]][i-];
         for(i=pre[node];i;i=e[i].next){
             v=e[i].to;
             ]) continue ;
             anc[v][]=node;
             dis[v][]=e[i].w;
             dfs(v,deep+);
         }
     }
     int lca(int x,int y){
         ,i;
         if(dep[x]<dep[y]) swap(x,y);
         ;i>=;i--)
             if(dep[y]<=dep[anc[x][i]]) res+=dis[x][i],x=anc[x][i];
         if(x==y) return res;
         ;i>=;i--)
             if(anc[x][i]!=anc[y][i]) res+=dis[x][i]+dis[y][i],x=anc[x][i],y=anc[y][i];
         ]+dis[y][];
     }
     int main(){
         memset(pre,,sizeof pre);
         int i,u,v,w;
         n=read(); m=read();
         cnt=;
         ;i<n;i++){
             u=read(); v=read(); w=read();
             add(u,v,w);
         }
         dfs(,);
         while(m--){
             u=read(); v=read();
             printf("%d\n",lca(u,v));
         }
         ;
     }
 }
 int main(){
     int T=read();
     while(T--) lys::main();
     ;
 }
 inline int read(){
     ,ff=;
     char c=getchar();
     '){
         ;
         c=getchar();
     }
     +c-',c=getchar();
     return kk*ff;
 }