B-CF31C Schedule

题解

枚举每个课，统计该课与别的课的相交情况（0<=book[i]<=n-1）,以及用S累计所有课有多少重合，两者做比较，满足条件即为S-book[i]==0，删掉该课后

#include<bits/stdc++.h>
using namespace std;
int l[5005], r[5005], book[5005];
int main()
{
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> l[i] >> r[i];
	int s = 0;
	for(int i=1;i<=n;i++)
		for (int j = 1; j <i; j++)
		{
			if (l[i] < r[j] && l[j] < r[i])
			{
				book[i]++, book[j]++,s++;//book[i] 记录第i个课与多少课有相交，s表示所有课的相交数之和
			}
		}
	vector<int>res;
	for (int i = 1; i <= n; i++)
	{
		if (book[i] - s==0)res.push_back(i);//如果当前的课相交数恰好等于最大的相交数，那么说明取消掉这节课剩下的课是不会有相交的
	}
	cout << res.size() << endl;
	for (int i = 0; i < res.size(); i++)
		cout << res[i] << " ";
}