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

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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;
}

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

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原文地址：http://blog.csdn.net/lianai911/article/details/42301329