Nearest Common Ancestors(LCA)(最近公共祖先LCA)

题目链接：

http://poj.org/problem?id=1330

这个问题做法比较多，记录下自己的写法。
1，首先将数据存储在邻接表里，先将数据按照并查集存储，然后将叶子节点的深度全部神深搜出来，存储到深度数组中。
2，然后就是具体做法：
先判断两个数是否在一棵树的同一层上，若不是先调整到同一层上。然后将两个数据在并查集内，同时向上搜寻，直至fath[a]==fath[b];

#include<stdio.h>
#include<iostream>
using namespace std;
#include<cstring>
int head[1000101];
int fath[1000001];
int deep[1000001];
int js;
int root;
int n;
struct edge
{
    int data;
    int next;
} v[100001];
void add(int x,int y)
{
    v[js].next = head[x];
    v[js].data = y;
    head[x] = js;
    js++;
}
void input()
{
    cin>>n;
    memset(head,-1,sizeof head);
    memset(fath,-1,sizeof fath);
    for (int i = 1; i < n; i++)
    {
        int x, y;
        cin >> x >> y;
        add(x,y);
        fath[y] = x;
    }

}
void dfs(int x,int d)
{
    for (int i = head[x]; i != -1; i = v[i].next)
        dfs(v[i].data,d+1);
    deep[x] = d;
}
void predeal()
{
    for(int i=1; i<=n; i++)
        if (fath[i] == -1)
        {
            root = i;
            break;
        }
    dfs(root,0);
}
void solve()
{
    int x, y;
    cin >> x >> y;
    if (deep[x] < deep[y])
    {
        swap(x,y);
    }
    while (deep[x] > deep[y])
        x = fath[x];
    while (x != y)
    {
        x = fath[x];
        y = fath[y];
    }
    printf("%d\n",x);
}
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        input();
        predeal();
        solve();
    }


}

标签：Nearest,head,Ancestors,int,void,fath,deep,js,LCA
来源： https://blog.csdn.net/weixin_46574282/article/details/121293745