【HDOJ2586】【Tarjan离线求LCA】

http://acm.hdu.edu.cn/showproblem.php?pid=2586

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21250    Accepted Submission(s): 8368

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

题目大意：给一个有权树，有Q次询问，求任意两点的距离。

题目分析：在树中有一个性质：任意两点的距离 ANS = dist[ u ] + dist[ v ] - dist[ lca(u,v)  ]，其中dist[ I ]是根到 I 的 距离 ，由于需要多次询问，所以使用Tarjan离线求LCA比较快

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int maxn=40005;
struct edge{
    int to;
    int len;
};
vector<struct edge>G[maxn];
vector<int>qury[maxn];
vector<int>num[maxn];
int dis[maxn],vis[maxn],ans[maxn],fa[maxn];
void init()
{
    memset(dis,0,sizeof(dis));
    memset(vis,0,sizeof(vis));
    for(int i = 0 ; i < 40005 ; i++)
    {
        G[i].clear();
        qury[i].clear();
        num[i].clear();
        fa[i]=i;
    }
}
int find(int x)
{
    int xx=x;
    while(fa[x]!=x)
    {
        x=fa[x];
    }
    while(fa[xx]!=x)
    {
        int t=fa[xx];
        fa[xx]=x;
        xx=t;
    }
    return x;
}
void Union(int x,int y)
{
    int xx=find(x);
    int yy=find(y);
    if(xx!=yy);
    fa[yy]=xx;//在完成子节点的Tarjan遍历之后，把子节点纳入父节点名下
}
void Tarjan(int u,int ll)
{
    vis[u]=1;
    dis[u]=ll;
    for(int i = 0 ; i< G[u].size() ;i++)
    {
        struct edge wqw=G[u][i];
        if(vis[wqw.to])continue;
        Tarjan(wqw.to,wqw.len+ll);
        Union(u,wqw.to);
    }
    for(int i = 0 ; i < qury[u].size() ; i++)
    {
        if(vis[qury[u][i]])
        {
            ans[num[u][i]]=dis[u]+dis[qury[u][i]]-2*dis[find(qury[u][i])];    
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i = 1 ; i < n ; i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            G[a].push_back((struct edge){b,c});
            G[b].push_back((struct edge){a,c});
        }
        for(int i = 0 ; i < m ; i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            qury[a].push_back(b);
            qury[b].push_back(a);
            num[a].push_back(i);
            num[b].push_back(i);
        }
        Tarjan(1,0);
        for(int i = 0 ; i < m ; i++)
        {
            printf("%d
",ans[i]);
        }
    }
    return 0; 
}