HDU2586 How far away ？【最近公共祖先】【Tarjan-LCA算法】

How far away ？

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

Total Submission(s): 6252    Accepted Submission(s): 2344

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

题目大意：一个村庄有N个房子和一些双向的路，人们总是喜欢问"A到B有多远呢"，一般是很难

回答的，毕竟有很多种答案。所幸，答案是唯一的，A到B总是有唯一的路径到达。第一行是T组

数据。每组数据第一行是N个房子和M条询问。接下来N-1行每行是u v w，表示从房子u到房子v

的距离是w。接下来是M行询问。每行是u v，表示询问房子u到房子v的距离，最后输出所有的询

问结果。

思路：整个村庄房子和路可看成一棵树，设根结点为房子1，询问u到房子v的距离，其实就是求u

到根结点的距离 + v到根结点的距离 - 2*(u，v)最近公共祖先到根结点的距离。这道题和POJ1986

是一样的，可参考：http://blog.csdn.net/lianai911/article/details/42300301

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 80080;
const int MAXQ = 440;

struct EdgeNode
{
    int to;
    int next;
    int lca;
}Edges[MAXN],QEdges[MAXQ];

int Head[MAXN],QHead[MAXN],father[MAXN],Dist[MAXN];
bool vis[MAXN];

int find(int x)
{
    if(x != father[x])
        father[x] = find(father[x]);
    return father[x];
}

void LCA(int u)
{
    father[u] = u;
    vis[u] = true;
    for(int k = Head[u]; k != -1; k = Edges[k].next)
    {
        if(!vis[Edges[k].to])
        {
            Dist[Edges[k].to] = Dist[u] + Edges[k].lca;
            LCA(Edges[k].to);
            father[Edges[k].to] = u;
        }
    }
    for(int k = QHead[u]; k != -1; k = QEdges[k].next)
    {
        if(vis[QEdges[k].to])
        {
            QEdges[k].lca = Dist[u] + Dist[QEdges[k].to] - 2*Dist[find(QEdges[k].to)];
            QEdges[k^1].lca = QEdges[k].lca;
        }
    }
}

int main()
{
    int T,N,M,u,v,w;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&N,&M);
        memset(Dist,0,sizeof(Dist));
        memset(father,0,sizeof(father));
        memset(vis,false,sizeof(vis));
        memset(Edges,0,sizeof(Edges));
        memset(QEdges,0,sizeof(QEdges));
        memset(Head,-1,sizeof(Head));
        memset(QHead,-1,sizeof(QHead));
        int id = 0;
        for(int i = 0; i < N-1; ++i)
        {
            scanf("%d%d%d",&u,&v,&w);
            Edges[id].to = v;
            Edges[id].lca = w;
            Edges[id].next = Head[u];
            Head[u] = id++;
            Edges[id].to = u;
            Edges[id].lca = w;
            Edges[id].next = Head[v];
            Head[v] = id++;
        }
        int ip = 0;
        for(int i = 0; i < M; ++i)
        {
            scanf("%d%d",&u,&v);
            QEdges[ip].to = v;
            QEdges[ip].next = QHead[u];
            QHead[u] = ip++;
            QEdges[ip].to = u;
            QEdges[ip].next = QHead[v];
            QHead[v] = ip++;
        }
        LCA(1);
        for(int i = 0; i < ip; i += 2)
            printf("%d\n",QEdges[i].lca);
    }
    return 0;
}