• AISing Programming Contest 2021(AtCoder Beginner Contest 202)E


    // Problem: C - Made Up
    // Contest: AtCoder - AISing Programming Contest 2021(AtCoder Beginner Contest 202)
    // URL: https://atcoder.jp/contests/abc202/tasks/abc202_c
    // Memory Limit: 1024 MB
    // Time Limit: 2000 ms
    // 
    // Powered by CP Editor (https://cpeditor.org)
    #pragma GCC optimize(2)
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<ll, ll>PLL;
    typedef pair<int, int>PII;
    typedef pair<double, double>PDD;
    #define I_int ll
    inline ll read()
    {
        ll x = 0, f = 1;
        char ch = getchar();
        while(ch < '0' || ch > '9')
        {
            if(ch == '-')f = -1;
            ch = getchar();
        }
        while(ch >= '0' && ch <= '9')
        {
            x = x * 10 + ch - '0';
            ch = getchar();
        }
        return x * f;
    }
     
    inline void out_(ll x){
        if (x < 0) x = ~x + 1, putchar('-');
        if (x > 9) out_(x / 10);
        putchar(x % 10 + '0');
    }
     
    inline void write(ll x){
        if (x < 0) x = ~x + 1, putchar('-');
        if (x > 9) write(x / 10);
        putchar(x % 10 + '0');
        puts("");
    }
     
    #define read read()
    #define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
    #define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
    #define rep(i,a,b) for(int i=(a);i<=(b);i++)
    #define repp(i,a,b) for(int i=(a);i<(b);i++)
    #define per(i,a,b) for(int i=(a);i>=(b);i--)
    #define perr(i,a,b) for(int i=(a);i>(b);i--)
    ll ksm(ll a, ll b, ll p)
    {
        ll res = 1;
        while(b)
        {
            if(b & 1)res = res * a % p;
            a = a * a % p;
            b >>= 1;
        }
        return res;
    }
    const int inf = 0x3f3f3f3f;
    #define PI acos(-1)
    const int maxn=2e5+100;
     
    int n,q;
    vector<int>g[maxn];
    vector<int>dis[maxn];
    int in[maxn],out[maxn],tot;
     
    void dfs(int u,int fa,int tmpd){
        in[u]=++tot;
        dis[tmpd].push_back(in[u]);
        for(int i=0;i<g[u].size();i++){
            int j=g[u][i];
            if(j==fa) continue;
            dfs(j,u,tmpd+1);
        }
        out[u]=tot;
    }
     
    int main(){
        n=read;
        rep(i,2,n){
            int x=read;
            g[x].push_back(i);
            g[i].push_back(x);      
        }
        dfs(1,-1,0);
        /*rep(i,1,n){
            cout<<i<<"*******"<<in[i]<<"******"<<out[i]<<"
    ";
        }
        for(int i=0;i<=4;i++){
            for(int j=0;j<dis[i].size();j++){
                cout<<dis[i][j]<<"=====";
            }
            puts("");
        }
        */
        q=read;
        while(q--){
            int u=read,d=read;
            int res=upper_bound(dis[d].begin(),dis[d].end(),out[u])-dis[d].begin();
            int ans=lower_bound(dis[d].begin(),dis[d].end(),in[u])-dis[d].begin();
            //cout<<res-ans<<endl;
            printf("%d
    ",res-ans);
        }
        return 0;
    } 
     
    
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  • 原文地址:https://www.cnblogs.com/OvOq/p/15049448.html
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