CodeForces 845C Two TVs (模拟)

C. Two TVs

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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).

Examples

input

3
1 2
2 3
4 5

output

YES

input

4
1 2
2 3
2 3
1 2

output

NO

一开始想错了，没有注意到这是两台电脑而只是看到不能在同一台电脑上看到两个节目的这个条件忽略了这是两台电脑。

思路：记录两台电视的使用情况以及当前播放节目的结束时间，然后模拟一下就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
int l,r;
}q[300000];
bool cmp(node a,node b)
{
if(a.l==b.l)
return a.r<b.r;
return a.l<b.l;
}
struct mach
{
int flag,r;
}tv[2];
int main()
{
int n;
while(~scanf("%d",&n))
{
tv[0].flag=0,tv[1].flag=0;
int f=1;
for(int i=0;i<n;i++)
scanf("%d%d",&q[i].l,&q[i].r);
sort(q,q+n,cmp);
for(int i=0;i<n;i++)
{
int time=q[i].l;
int flag=0;
if(!tv[0].flag)
{
tv[0].flag=1;
tv[0].r=q[i].r;
flag=1;
}
else if(tv[0].flag)
{
if(time>tv[0].r)
{
tv[0].r=q[i].r;
flag=1;
}
}
if(!tv[1].flag&&!flag)
{
tv[1].flag=1;
tv[1].r=q[i].r;
flag=1;
}
else if(tv[1].flag&&!flag)
{
if(time>tv[1].r)
{
tv[1].r=q[i].r;
flag=1;
}
else
{
f=0;
break;
}
}
if(!flag)
{
f=0;
break;
}
}
if(f)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}