hdu 2583 How far away ？ 离线算法 带权求最近距离

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28961    Accepted Submission(s): 11639

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n个节点的带权树，m次询问，求两点之间的距离
第一行输入t,表示多个测试样例
第二行输入n,m，表示n个节点和m次询问
接下来n-1行，每行输入x,y,w，表示x到y的距离为w
最后m行每行输入两个数xx,yy，询问xx到yy的最短距离为多少

 

这题给人的感觉就是最短路可以做，但是注意数据范围40000，开不了40000的二维数组，所以直接gg

但其实就是简单的最近公共祖先问题，找到x,y两点最近的公共祖先节点z[i]，dis[x]+dis[y]-2*dis[z[i]]就是答案

 

 

#include<iostream>
#include<string.h>
#include<vector>
using namespace std;
struct node
{
    int id;
    int len;
}p;
vector<node>mp[];//结构体动态数组
int dis[],xx[],yy[],vis[],z[],fa[];//dis是存距离，xx,yy存询问的点，z[i]表示i的LCA是点z[i]，fa是存父亲节点
int t,n,m;
void add(int a,int b,int c)//加边建树
{
    p.id=b;
    p.len=c;
    mp[a].push_back(p);
}
void init()//初始化
{
    for(int i=;i<=n;i++)
    {
        fa[i]=i;
        mp[i].clear();
        dis[i]=;
        vis[i]=;
        z[i]=;
    }
}
int find(int x)
{
    if(fa[x]!=x)
        return fa[x]=find(fa[x]);
    return fa[x];

}

void join(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x!=y)
        fa[y]=x;
}
void tarjan(int x)
{
    vis[x]=;
    for(int i=;i<mp[x].size();i++)
    {
        int y=mp[x][i].id;
        if(!vis[y])//未遍历过
        {
            dis[y]=dis[x]+mp[x][i].len;//更新距离
            tarjan(y);//遍历下一个点
            join(x,y);//合并
        }

    }
    for(int i=;i<=m;i++)
    {
        if(xx[i]==x&&vis[yy[i]])
            z[i]==find(yy[i]);
        if(yy[i]==x&&vis[xx[i]])
            z[i]=find(xx[i]);
    }

}
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        int a,b,c;
        init();
        for(int i=;i<n;i++)//描述边
        {
            cin>>a>>b>>c;
            add(a,b,c);
            add(b,a,c);
        }
        for(int i=;i<=m;i++)//询问
        {
            cin>>xx[i]>>yy[i];
        }
        tarjan();
        for(int i=;i<=m;i++)
            cout<<dis[xx[i]]+dis[yy[i]]-*dis[z[i]]<<endl;
    }
    return ;
}