How far away ？（LCA）dfs和倍增模版

How far away ？

 Tarjan

http://www.cnblogs.com/caiyishuai/p/8572859.html

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20491    Accepted Submission(s): 8010

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

 

 

Source

ECJTU 2009 Spring Contest

 

 

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这一道题目意思是说，村庄之间有路可达，给你N个节点，N-1条路，然后M组查询，查询两个节点之间的距离。

N个节点N-1条边，那么就符合树的定义。所以题目给的就是一个树，就是求树上两个节点的距离。


这一道题目可以和LCA联系起来，求两个节点（a和b）的距离。两个节点必然由一个公共点连接起来，这个点就是LCA（最近公共祖先c）

那么求距离就可以转换为a到根节点的距离+b到根节点的距离—c到根节点的距离—c到根节点的距离。

 

 

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>  
#include<functional>  
#define N 100000+10  
using namespace std;  
int n,m;  
struct node  
{  
    int to,next,cost;  
}e[N];  
int cnt;  
int fa[20][N];  
int head[N],depth[N],dis[N];  
void init()  
{  
    memset(head,-1,sizeof head);  
    memset(depth,0,sizeof depth);  
    memset(dis,0,sizeof dis);  
    cnt=0;  
}  
void addedge(int u,int v,int w)//建图过程，建双向边  
{  
    e[cnt].to=v;  
    e[cnt].cost=w;  
    e[cnt].next=head[u];  
    head[u]=cnt++;  
}

void DFS(int u,int f)//遍历树  
{  
    fa[0][u]=f;  
    for(int i=head[u];~i;i=e[i].next)//遍历所有相连的边  
    {  
        int To=e[i].to;  
        if(To!=f)//去掉以后MLE，可能是递归求的过程中太多临时变量  
        {//建树过程建双向边，会出现to=f的情况，去掉以后会陷入无限递归中  
            dis[To]=dis[u]+e[i].cost;//更新距离  
            depth[To]=depth[u]+1;//更新深度  
            DFS(To,u);  
        }  
    }  
  
}  

void solve()  
{  
    depth[1]=1;//题目给的是一个树  
    dis[1]=0;//无论怎么样的树，都可以把1视为根节点  
    DFS(1,-1);  
    for(int i=1;i<20;i++)//树上倍增  
        for(int j=1;j<=n;j++)  
            fa[i][j]=fa[i-1][fa[i-1][j] ];  
}  


int LCA(int u,int v)//求最近公共祖先  
{  
    if(depth[u]>depth[v])//保证V的深度比较大  
        swap(u,v);  
    for(int i=0;i<20;i++)//倍增到深度相同  
        if((depth[v]-depth[u])>>i&1)//二进制特性，一定能跳到深度相同  
            v=fa[i][v];  
    if(u==v)  
        return u;  
    for(int i=19;i>=0;i--)//两者同时倍增  
    {  
        if(fa[i][u]!=fa[i][v])  
        {  
            u=fa[i][u];  
            v=fa[i][v];  
        }  
    }  
    return fa[0][v];  
}  
int main()  
{  
    int i,t;  
    int a,b,c;  
    while(scanf("%d",&t)!=EOF)  
    {  
  
        while(t--)  
        {  
            init();  
            scanf("%d%d",&n,&m);  
            for(i=1;i<n;i++)  
            {  
                scanf("%d%d%d",&a,&b,&c);  
                addedge(a,b,c);  
                addedge(b,a,c);  
            }  
            solve();  
            for(i=1;i<=m;i++)  
            {  
                scanf("%d%d",&a,&b);  
                int ans=dis[a]+dis[b]-2*dis[LCA(a,b)];  
                printf("%d\n",ans);  
            }  
        }  
        return 0;  
    }  
}