2022牛客多校第三场 A Ancestor

题目链接

Ancestor

题目大意

给了我们两棵树$A$和$B$，然后给了我们$k$个点，问我们分别在$A$和$B$删去其中一个点，使得剩余点在$A$树中的最近公共祖先的权值大于剩余点在$B$树中的最近公共祖先的权值。

题解

首先，预处理出来，这$k$个点的前缀最近公共祖先和后缀最近公共祖先，然后，分别枚举$k$个点，将该点去掉的最近公共祖先就是$lca(pre[i-1],back[i+1])$，如此便可进行判断。

代码

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define IOS ios::sync_with_stdio(false); cin.tie(0);cout.tie(0);
#define x first
#define y second
#define int long long
#define endl '\n' 
const int inf = 1e9 + 10;
const int maxn = 100010, M = 200010;
const int mod = 1e9 + 7;
typedef pair<int,int> PII;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
int key[maxn];
int n, k;
struct Tree
{
    
    int h[maxn], e[M], ne[M], idx;
    int v[maxn];
    int pre[maxn], back[maxn];
    int fa[maxn][20];
    int depth[maxn];
    Tree()
    {
    
        memset(h, -1, sizeof h);
    }
    void add(int a, int b)
    {
    
        e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
    }
    void bfs(int root)
    {
    
        memset(depth, 0x3f, sizeof depth);
        depth[0] = 0, depth[root] = 1;
        queue<int> q;
        q.push(root);
        while(q.size())
        {
    
            auto t = q.front();
            q.pop();
            for(int i = h[t]; ~i; i = ne[i]) {
    
                int j = e[i];
                if(depth[j] > depth[t] + 1) {
    
                    depth[j] = depth[t] + 1;
                    q.push(j);
                    fa[j][0] = t;
                    for(int k = 1; k <= 18; k ++) {
    
                        fa[j][k] = fa[fa[j][k - 1]][k - 1];
                    }
                }
            }
        }
    }
    int lca(int a, int b)
    {
    
        if(a == 0) return b;
        if(b == 0) return a;
        if(depth[a] < depth[b]) swap(a, b);
        for(int k = 18; k >= 0; k --){
    
            if(depth[fa[a][k]] >= depth[b]) {
    
                a = fa[a][k];
            }
        }
        if(a == b) return a;
        for(int k = 18; k >= 0; k --){
    
            if(fa[a][k] != fa[b][k]) {
    
                a = fa[a][k];
                b = fa[b][k];
            }
        }
        return fa[a][0];
    }
    void init()
    {
    
        bfs(1);
        pre[1] = key[1], back[k] = key[k];
        for(int i = 2; i <= k; i ++) pre[i] = lca(pre[i - 1], key[i]);
        for(int i = k - 1; i >= 1; i --) back[i] = lca(back[i + 1], key[i]);
    }
    int calc(int x)
    {
    
        return v[lca(pre[x - 1], back[x + 1])];
    }
}ta, tb;
signed main()
{
    
    IOS;
    cin >> n >> k;
    for(int i = 1; i <= k; i ++) cin >> key[i];
    for(int i = 1; i <= n; i ++) cin >> ta.v[i];
    for(int i = 2; i <= n; i ++){
    
        int x; cin >> x;
        ta.add(x, i);
    }
    for(int i = 1; i <= n; i ++) cin >> tb.v[i];
    for(int i = 2; i <= n; i ++){
    
        int x; cin >> x;
        tb.add(x, i);
    }
    ta.init();
    tb.init();
    int res = 0;
    for(int i = 1; i <= k; i ++){
    
        int t1 = ta.calc(i), t2 = tb.calc(i);
        if(t1 > t2) res ++;
    }
    cout << res << endl;
    return 0;
}