LCA在线算法（hdu2586）

hdu2586

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4183    Accepted Submission(s): 1598

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
  
    

分析：

LCA最近公共祖先其实就是利用树形结构求两点间的最短路；一般题目说有n个点，且仅有n-1条边，则优先考虑LCA算法（离线或在线）

程序：

#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#define M 40009
#include"string"
#include"map"
#include"iostream"
using namespace std;
struct st
{
    int u,v,next,w;
}edge[M];
int head[M],f[M],rank[M],use[M],dis[M],t;
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 dfs(int u)
{
    int i;
    use[u]=1;
    for(i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!use[v])
        {
            rank[v]=rank[u]+1;
            dis[v]=dis[u]+edge[i].w;
            dfs(v);
        }
    }
}
int LCA(int u,int v)
{
    if(u==v)
        return u;
    else if(rank[u]>rank[v])
        return LCA(f[u],v);
    else
        return LCA(u,f[v]);
}
int main()
{
    int T,i,a,b,c,m,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            f[i]=i;
        init();
        for(i=1;i<n;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            f[b]=a;
        }
        memset(dis,0,sizeof(dis));
        memset(rank,0,sizeof(rank));
        memset(use,0,sizeof(use));
        for(i=1;i<=n;i++)
            if(f[i]==i)
            dfs(i);
        while(m--)
        {
            scanf("%d%d",&a,&b);
            int ans=LCA(a,b);
            printf("%d\n",dis[a]+dis[b]-2*dis[ans]);
        }
    }
    return 0;
}