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Blood Cousins dsu on tree + k祖先查询

作者:互联网

题意

给出一棵家谱树,定义从 u 点向上走 k 步到达的节点为 u 的 k-ancestor。多次查询,给出 u k,问有多少个与 u 具有相同 k-ancestor 的节点。

分析

这个问题我们可以离线去查询

首先我们先把每一个查询节点的k-ancesto处理出来,然后就把这个问题转化成求节点x的子树中,比他深度大k的节点有多少,就是一个树上静态统计问题了,可以用dsu on tree来进行处理

代码

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <queue>
#include <cstring>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define _CRT_SECURE_NO_WARNINGS
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline")
#pragma GCC option("arch=native","tune=native","no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
int h[N],ne[N],e[N],idx;
int dep[N];
int cnt[N];
int son[N],Size[N],Son;
int root[N];
int f[N][21];
int ans[N];
vector<PII> q[N];
int n,m;

void add(int x,int y){
    ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}

void dfs(int u,int fa){
    Size[u] = 1;
    f[u][0] = fa;
    for(int i = 1;i <= 20;i++)
        f[u][i] = f[f[u][i - 1]][i - 1];
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        dep[j] = dep[u] + 1;
        dfs(j,u);
        Size[u] += Size[j];
        if(Size[son[u]] < Size[j]) son[u] = j;
    }
}

int getdep(int x,int d){
    int p = dep[x] - d;
    for(int i = 20;~i;i--)
        if(dep[f[x][i]] > p) x = f[x][i];
    return f[x][0];
}

void cal(int u,int st){
    cnt[dep[u]] += st;
    for(int i = h[u];~i;i = ne[i]){
        if(e[i] != Son)
            cal(e[i],st);
    }
}

void dsu(int u,bool st){
    for(int i = h[u];~i;i = ne[i]){
        if(e[i] != son[u]) dsu(e[i],0);
    }
    if(son[u]) dsu(son[u],1);
    Son = son[u];
    cal(u,1);
    Son = 0;
    for(auto t:q[u]){
        int id = t.first,d = t.second;
        ans[id] = cnt[dep[u] + d];
    }
    if(!st) cal(u,-1);
}

int main(){
    memset(h,-1,sizeof h);
    scanf("%d",&n);
    int ll = 0;
    for(int i = 1;i <= n;i++){
        int fa;scanf("%d",&fa);
        add(fa,i);
    }
    dfs(0,0);
    scanf("%d",&m);
    for(int i = 0;i < m;i++){
        int x,y,z;
        scanf("%d%d",&x,&y);
        z = getdep(x,y);
        if(!z){
            ans[i] = 1;
            continue;
        }
        q[z].push_back({i,y});
    }
    dsu(0,1);
    for(int i = 0;i < m;i++) printf("%d ",ans[i] - 1);
    return 0;
}

/**
*  ┏┓   ┏┓+ +
* ┏┛┻━━━┛┻┓ + +
* ┃       ┃
* ┃   ━   ┃ ++ + + +
*  ████━████+
*  ◥██◤ ◥██◤ +
* ┃   ┻   ┃
* ┃       ┃ + +
* ┗━┓   ┏━┛
*   ┃   ┃ + + + +Code is far away from  
*   ┃   ┃ + bug with the animal protecting
*   ┃    ┗━━━┓ 神兽保佑,代码无bug 
*   ┃        ┣┓
*    ┃        ┏┛
*     ┗┓┓┏━┳┓┏┛ + + + +
*    ┃┫┫ ┃┫┫
*    ┗┻┛ ┗┻┛+ + + +
*/
/**
*  ┏┓   ┏┓+ +
* ┏┛┻━━━┛┻┓ + +
* ┃       ┃
* ┃   ━   ┃ ++ + + +
*  ████━████+
*  ◥██◤ ◥██◤ +
* ┃   ┻   ┃
* ┃       ┃ + +
* ┗━┓   ┏━┛
*   ┃   ┃ + + + +Code is far away from  
*   ┃   ┃ + bug with the animal protecting
*   ┃    ┗━━━┓ 神兽保佑,代码无bug 
*   ┃        ┣┓
*    ┃        ┏┛
*     ┗┓┓┏━┳┓┏┛ + + + +
*    ┃┫┫ ┃┫┫
*    ┗┻┛ ┗┻┛+ + + +
*/



标签:int,Cousins,dsu,tree,son,dep,include,Size
来源: https://blog.csdn.net/tlyzxc/article/details/110733553