hdu2586（LCA最近公共祖先）

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3653    Accepted Submission(s): 1379

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

 

题目分析：题目中说有n个房子，有n-1条道路，并且都是联通的，典型的树状图，要求任意两点的距离，可以用线段树做，此处不提，用最短路

n的范围是4000，用一般最短路的方法一定会超时，因此利用树状结构的特点，最好用LCA；

程序：

#include"stdio.h"
#include"string.h"
#include"iostream"
#include"queue"
#include"map"
using namespace std;
#define M 40005
int dis[M],pre[M],head[M],t,sum,use[M],rank[M];
struct st
{
    int u,v,w,next;
}edge[M*3];
void init()
{
    t=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
    edge[t].u=u;
    edge[t].v=v;
    edge[t].w=w;
    edge[t].next=head[u];
    head[u]=t++;
}
void bfs(int s)
{
    queue<int>q;
    memset(use,0,sizeof(use));
    memset(rank,0,sizeof(rank));
    use[s]=1;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v;
            if(!use[v])
            {
                use[v]=1;
                pre[v]=u;//记录父节点；
                rank[v]=rank[u]+1;//记录层数
                dis[v]=edge[i].w;//记录子节点到父节点的距离
                q.push(v);
            }
        }
    }
}
void targan(int a,int b)
{
    if(a==b)
    return;
    else if(rank[a]>rank[b])
    {
        sum+=dis[a];
        targan(pre[a],b);
    }
    else
    {
        sum+=dis[b];
        targan(a,pre[b]);
    }
}//用深搜的方法求最近公共祖先；sum记录路径长度
int main()
{
    int w,a,b,c,i,m,n;
    scanf("%d",&w);
    while(w--)
    {
        init();
        scanf("%d%d",&n,&m);
        for(i=1;i<n;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        bfs(1);
        while(m--)
        {
            scanf("%d%d",&a,&b);
            sum=0;
            targan(a,b);
            printf("%d\n",sum);
        }
    }
}