codeforce_845c_Two TVs

版权声明： https://blog.csdn.net/jianbagengmu/article/details/79797521

Two TVs

Polycarp is a great fan of television.

He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment li and ends at moment ri.

Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can’t watch them on a single TV.

Polycarp wants to check out all n shows. Are two TVs enough to do so?

Input
The first line contains one integer n (1 ≤ n ≤ 2·105) — the number of shows.

Each of the next n lines contains two integers li and ri (0 ≤ li < ri ≤ 109) — starting and ending time of i-th show.

Output
If Polycarp is able to check out all the shows using only two TVs then print “YES” (without quotes). Otherwise, print “NO” (without quotes).

Sample Input
Input
3
1 2
2 3
4 5
Output
YES
Input
4
1 2
2 3
2 3
1 2
Output
NO
Source
Educational Codeforces Round 27
Language: Theme:

1
​
Share Code? Yes No Submit

mean ：
输入n，接下来有n个电视节目，给出开始和结束时间；

有两台tv 可以同时使用，但是当一个节目结束，另一个节目立刻开始是没法在同一个电视操作的

ans ;
模拟

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+99;
struct node
{
    int u,v;
}num[N];
bool cmp(node a,node b)
{
    if(a.u!=b.u) return a.u<b.u;
    return a.v<b.v;
}
int main()
{
    int n,i,j;
    while(cin>>n)
    {
        for(i=0;i<n;i++)
            scanf("%d%d",&num[i].u,&num[i].v);
        sort(num,num+n,cmp);
        int stu[2]={-1};
        stu[1]=-1,stu[2]=-1;

        int f=0;
        for(i=0;i<n;i++)
        {
            int f1=0;
            int f2=0;
            if(num[i].u>stu[0])
            {
                f1=1;
                stu[0]=num[i].v;
                continue;
            }
            if(num[i].u>stu[1])
            {
                f2=1;
                stu[1]=num[i].v;
                continue;
            }
            if(f1+f2==0)
            {
                f=1;
                break;
            }
        }
        if(f)
            puts("NO");
        else puts("YES");
    }
}