样例

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算法1

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define x first
#define y second
using namespace std;
typedef pair<int,int> PII;
const int N = 100010;
int score[N],last[N]; 
bool st[N];
int n,m,T;
int res;
PII order[N];
int main()
{
    cin >> n >> m >> T;
    for(int i = 0; i < m; i++) scanf("%d",&order[i].x,&order[i].y);
    sort(order,order + m);

    for(int i = 0; i < m;)
    {
        int j = i;
        while(j < m && order[j] == order[i]) j++;

        int t = order[i].x,id = order[i].y,cnt = j - i;
        i = j;

        score[id] -= t - last[id] - 1;
        if(score[id] < 0) score[id] = 0;
        if(score[id] <= 3) st[id] = false;

        score[id] += cnt * 2;
        if(score[id] > 5) st[id] = true;

        last[id] = t;
    }

    for(int i = 1; i <= n; i++)
    {
        if(last[i] < T)
        {
            score[i] -= T - last[i];
            if(score[i] <= 3) st[i] = false;
        }
    }

    for(int i = 1; i <= n; i++)
    {
        res += st[i];
    }
    cout << res << endl;

    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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