B. Turn the Rectangles
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
There are nn rectangles in a row. You can either turn each rectangle by 9090 degrees or leave it as it is. If you turn a rectangle, its width will be height, and its height will be width. Notice that you can turn any number of rectangles, you also can turn all or none of them. You can not change the order of the rectangles.
Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).
Input
The first line contains a single integer nn (1≤n≤1051≤n≤105) — the number of rectangles.
Each of the next nn lines contains two integers wiwi and hihi (1≤wi,hi≤1091≤wi,hi≤109) — the width and the height of the ii-th rectangle.
Output
Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
input
Copy
3 3 4 4 6 3 5
output
Copy
YES
input
Copy
2 3 4 5 5
output
Copy
NO
Note
In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].
In the second test, there is no way the second rectangle will be not higher than the first one.
题意：给了n个矩形的长和宽，随意交换长宽，让矩形的长或者宽形成非递增排列。
题解：贪心思想 第一次取最大的，后面每次取的时候先拿最大的和前面一个比较，<＝就存进去，如果不行就拿小的看是<=前面那个存进去，否则输出no
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,temp,a,b;
cin>>n;
cin>>a>>b;
temp=max(a,b);
for(int i=1; i<n; i++)
{
cin>>a>>b;//采取贪心策略，先取大的，看是否小于等于前一位，否，则取小的，再否，不符条件
if (temp>=max(a,b)) temp=max(a,b);
else if (temp>=min(a,b)) temp=min(a,b);
else
{
printf("NO");
return 0;
}
}
printf("YES");
return 0;
}