洛谷 P3019 [USACO11MAR]会见点Meeting Place-优快云博客

本文介绍了一个经典的LCA算法实现案例，该算法用于解决在给定的树状结构中找到两个节点的最近公共祖先的问题。文章提供了完整的代码示例，并解释了如何通过深度优先搜索预处理树结构以高效地查询最近公共祖先。

题目背景

征求翻译。如果你能提供翻译或者题意简述，请直接发讨论，感谢你的贡献。

题目描述

Bessie and Jonell are great friends. Since Farmer John scrambles where the cows graze every day, they are sometimes quite far from each other and can't talk.

The pastures and paths on FJ's farm form a 'tree' structure. Each pasture has exactly one distinct path to any other pasture, and each pasture (except pasture #1, the 'root') also has a single parent node.

Bessie and Jonell have decided that they will always meet at the closest pasture that that is both an ancestor of Jonell's pasture and of Bessie's pasture.

FJ created a map of his N (1 <= N <= 1,000) pastures (conveniently numbered 1..N) that tells the parent P_i (1 <= P_i <= N) of each pasture except pasture 1, which has no parent.

FJ has released his daily grazing schedule for the next M (1 <= M <= 1,000) days, so Bessie and Jonell are deciding where they should meet each day for gossip. On day k, Bessie is in pasture B_k (1 <= B_k <= N) and Jonell is in pasture J_k (1 <= J_k <= N).

Given a map and schedule, help Bessie and Jonell find their meeting places.

Consider, for example, the following farm layout:

                            Pasture      Parent Pasture
             [1]           ---------    ----------------
            / | \              1              ---
           /  |  \             2               1 
         [2] [3] [6] 3 1 / | \ 4 2 / | \ 5 8 [4] [8] [9] 6 1 / \ 7 8 / \ 8 6 [5] [7] 9 6 Here are the meeting places that Bessie and Jonell would choose given a six day schedule of their initial grazing locations: Bessie Jonell Meeting Place -------- -------- --------------- 2 7 1 4 2 2 1 1 1 4 1 1 7 5 8 9 5 6

输入输出格式

输入格式：

Line 1: Two space-separated integers: N and M
Lines 2..N: Line i contains a single integer that describes the parent of pasture i: P_i
Lines N+1..N+M: Line k+N describes Bessie and Jonell's respective pastures with two space-separated integers: B_k and J_k

输出格式：

Lines 1..M: Line j contains the meeting place Bessie and Jonell would use for line j+N of the input

输入输出样例

输入样例#1：

输出样例#1：

lca模板题

屠龙宝刀点击就送

#include <vector>
#include <cstdio>
#define N 1005

using std::vector;
int n,m,dad[N][25],dep[N],siz[N];
vector<int>G[N];
void dfs(int x)
{
    dep[x]=dep[dad[x][0]]+1;
    for(int i=0;dad[x][i];++i)
    dad[x][i+1]=dad[dad[x][i]][i];
    for(int i=0;i<G[x].size();++i)
    {
        int v=G[x][i];
        if(dad[x][0]!=v)
        {
            dad[v][0]=x;
            dfs(v);
        }
    }
}
inline void swap(int &m,int &n)
{
    int tmp=n;
    n=m;
    m=tmp;
}
int lca(int x,int y)
{
    if(dep[x]>dep[y]) swap(x,y);
    for(int i=20;i>=0;--i)
    if(dep[dad[y][i]]>=dep[x]) y=dad[y][i];
    if(x==y) return x;
    for(int i=20;i>=0;--i)
    if(dad[x][i]!=dad[y][i]) x=dad[x][i],y=dad[y][i];
    return dad[x][0];
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int x,i=2;i<=n;++i)
    {
        scanf("%d",&x);
        G[x].push_back(i);
        G[i].push_back(x);  
    }
    dfs(1);
    for(int x,y;m--;)
    {
        scanf("%d%d",&x,&y);
        printf("%d\n",lca(x,y));
    }
    return 0;
}