CodeForces 1883D In Love

合集 - Codeforces(21)

1.CodeForces 1992E Novice's Mistake2024-07-162.CodeForces 1935A Entertainment in MAC2024-06-163.CodeForces 1992C Gorilla and Permutation2024-07-164.CodeForces 1992D Test of Love2024-07-165.CodeForces 1992B Angry Monk2024-07-166.CodeForces 1992A Only Pluses2024-07-167.CodeForces 1983B Corner Twist2024-07-168.CodeForces 1983A Array Divisibility2024-07-169.CodeForces 1983C Have Your Cake and Eat It Too2024-07-1610.CodeForces 1883B Chemistry2024-07-2611.CodeForces 1883A Morning2024-07-2612.CodeForces 1883C Raspberries2024-07-28

13.CodeForces 1883D In Love2024-07-28

14.CodeForces 1883E Look Back2024-07-2815.CodeForces 1883F You Are So Beautiful2024-07-2816.CodeForces 1883G1 Dances (Easy version)2024-07-2817.CodeForces 908B New Year and Buggy Bot2024-08-0118.CodeForces 908C New Year and Curling2024-08-0119.CodeForces 1132B Discounts2024-08-0120.CodeForces 1619D New Year's Problem2024-08-0121.Codeforces Round 964 (Div. 4)2024-08-07

收起

题目链接：CodeForces 1883D【In Love】

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思路

       求能否找出两个区间不相交，所以将得到的区间先按区间左端点的大小从小到大排列，再按区间右端点的大小从小到大排列，此时取出最小的右端点和最大的左端点，若右端点在左端点左侧，则存在两个不相交的区间。由于需要动态操作增加减少区间，所以可以使用map优化。

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代码

#include <bits/stdc++.h>
using namespace std;
 
multiset<int> ml, mr;
 
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        char op[2];
        scanf("%s", op);
        int l, r;
        scanf("%d%d", &l, &r);
        if(*op == '-')
        {
            ml.erase(ml.find(l));
            mr.erase(mr.find(r));
        }
        else
        {
            ml.insert(l);
            mr.insert(r);
        }
        if(!ml.size())
        {
            cout << "NO" << endl;
        }
        else if(*(prev(ml.end())) > *mr.begin())
        {
            cout << "YES" << endl;
        }
        else
        {
            cout << "NO" << endl;
        }
    }
    return 0;
}