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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#486995#7736. Red Black TreeCSQTL 4ms19268kbC++142.1kb2024-07-22 14:59:182024-07-22 14:59:18

Judging History

你现在查看的是最新测评结果

  • [2024-07-22 14:59:18]
  • 评测
  • 测评结果:TL
  • 用时:4ms
  • 内存:19268kb
  • [2024-07-22 14:59:18]
  • 提交

answer

#pragma GCC optimize("Ofast") 
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define sz(a) (int)(a.size())
#define all(a) a.begin(),a.end()
#define lb lower_bound
#define ub upper_bound
#define owo ios_base::sync_with_stdio(0);cin.tie(0);
#define debug(...) fprintf(stderr, __VA_ARGS__),fflush(stderr)
#define time__(d) for(long blockTime = 0; (blockTime == 0 ? (blockTime=clock()) != 0 : false);\
debug("%s time : %.4fs\n", d, (double)(clock() - blockTime) / CLOCKS_PER_SEC))
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> PII;
typedef pair<int,int> pii;
typedef vector<vector<int>> vii;
typedef vector<vector<ll>> VII;
ll gcd(ll a,ll b){if(!b)return a;else return gcd(b,a%b);}
const int MAXN = 2e5+5;
vector<int>adj[MAXN];
string s;
multiset<int>dp[MAXN];
int best[MAXN],sum[MAXN],ans[MAXN];
void dfs(int v){
    dp[v].clear();
    sum[v] = best[v] = 0;
    for(int x:adj[v]){
        dfs(x);
        if(dp[v].empty()){
            best[v] = best[x];
            sum[v] = sum[x];
            dp[v] = dp[x];
        }else{
            if(sz(dp[v]) > sz(dp[x])){
                swap(dp[v],dp[x]);
                swap(best[v],best[x]);
            }
            vector<int>c;
            auto it = dp[x].begin();
            sum[v] = 0;
            for(int val:dp[v]){
                c.pb(val+*it);
                if(val + *it <= 0)sum[v] += val + *it;
                it++;
            }
            best[v]+=best[x];
            dp[v] = multiset<int>(c.begin(),c.end());
        }
    }
    if(s[v]=='0')dp[v].insert(1);
    else {
        dp[v].insert(-1);
        best[v]++;
        sum[v]--;
    }
    ans[v] = best[v] + sum[v];
}
int main()
{
    owo
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        cin>>s;
        for(int i=1;i<n;i++){
            int p;
            cin>>p;
            adj[p-1].pb(i);
        }
        dfs(0);
        for(int i=0;i<n;i++)cout<<ans[i]<<" ";
        cout<<'\n';
        for(int i=0;i<n;i++)adj[i].clear();
    }


}

詳細信息

Test #1:

score: 100
Accepted
time: 4ms
memory: 19268kb

input:

2
9
101011110
1 1 3 3 3 6 2 2
4
1011
1 1 3

output:

4 1 2 0 0 0 0 0 0 
2 0 0 0 

result:

ok 2 lines

Test #2:

score: -100
Time Limit Exceeded

input:

6107
12
000000001000
1 2 3 2 5 4 4 7 3 8 11
19
1100111101111011110
1 2 1 1 4 5 2 4 3 2 2 7 10 2 11 3 15 5
7
0111110
1 1 2 2 1 5
3
000
1 1
7
1000011
1 2 3 3 5 4
7
0001001
1 1 1 3 5 3
8
00111000
1 1 3 2 5 2 7
11
11111110111
1 1 1 4 5 4 5 2 5 1
15
110101101000010
1 2 3 2 1 5 2 5 6 5 8 7 9 14
10
0101000...

output:

1 1 1 1 0 0 0 0 0 0 0 0 
6 2 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
1 0 0 0 0 0 0 
0 0 0 
0 0 0 0 0 0 0 
2 0 1 0 0 0 0 
2 1 0 0 0 0 0 0 
4 0 0 2 1 0 0 0 0 0 0 
4 3 0 0 2 0 0 0 0 0 0 0 0 0 0 
2 0 1 0 0 0 0 0 0 0 
6 5 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 
1 1 0 0 0 
5 1 0 1 0 0 0 0 0 0 0 0 0 0 
1 0 0 0 0 0 0 0 0...

result: