LCA最近公共祖先知识点整理

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

3 2

1 2 10

3 1 15

1 2

2 3

2 2

1 2 100

1 2

2 1

Sample Output

100

解题思路：Tarjan算法求树上任意两点的最短路。

AC代码：

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long LL;

 const int maxn=;

 const int maxv=;

 int T,n,m,x,y,z,fa[maxn],dist[maxn],ans[maxv];bool vis[maxn];

 vector<pair<int,int> > tree[maxn],query[maxn];

 void init(){

     for(int i=;i<=n;++i)fa[i]=i,tree[i].clear(),query[i].clear();

     memset(vis,false,sizeof(vis));

     memset(dist,,sizeof(dist));

     memset(ans,,sizeof(ans));

 }

 int query_father(int x){

     int per=x,tmp;

     while(fa[per]!=per)per=fa[per];

     while(x!=per){tmp=fa[x];fa[x]=per;x=tmp;}//路径压缩

     return x;

 }

 void unite(int x,int y){

     x=query_father(x),y=query_father(y);

     if(x!=y)fa[y]=x;

 }

 void Tarjan_LCA(int cur,int val){

     vis[cur]=true;///先标记为true，这样如果同一支树上前面的祖先节点已访问，但是还没有归并到集合中去相当于为其本身，所以对下面计算两个节点的最近距离是没有影响的

     dist[cur]=val;

     for(size_t i=;i<tree[cur].size();++i){

         int v=tree[cur][i].first;

         int w=tree[cur][i].second;

         if(!vis[v]){

             Tarjan_LCA(v,val+w);///先序遍历

             unite(cur,v);///回退后续遍历归并集合

         }

     }

     for(size_t i=;i<query[cur].size();++i){///同时扫描与cur有关的点对查询

         int to=query[cur][i].first;

         int id=query[cur][i].second;

         if(vis[to])ans[id]=dist[cur]+dist[to]-*dist[query_father(to)];

     }

 }

 int main(){

     while(~scanf("%d",&T)){

         while(T--){

             scanf("%d%d",&n,&m);

             init();

             for(int i=;i<n;++i){

                 scanf("%d%d%d",&x,&y,&z);

                 tree[x].push_back(make_pair(y,z));

                 tree[y].push_back(make_pair(x,z));

             }

             for(int i=;i<=m;++i){

                 scanf("%d%d",&x,&y);

                 query[x].push_back(make_pair(y,i));

                 query[y].push_back(make_pair(x,i));

             }

             Tarjan_LCA(,);

             for(int i=;i<=m;++i)printf("%d\n",ans[i]);

         }

     }

     return ;

 }

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