cf MemSQL start[c]up Round 1 A Square and Rectangles

A. Square and Rectangles

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n rectangles. The corners of rectangles have integer coordinates and their edges are parallel to the Ox and Oy axes. The rectangles may touch each other, but they do not overlap (that is, there are no points that belong to the interior of more than one rectangle).

Your task is to determine if the rectangles form a square. In other words, determine if the set of points inside or on the border of at least one rectangle is precisely equal to the set of points inside or on the border of some square.

Input

The first line contains a single integer n (1 ≤ n ≤ 5). Next n lines contain four integers each, describing a single rectangle: x1, y1, x2, y2(0 ≤ x1 < x2 ≤ 31400, 0 ≤ y1 < y2 ≤ 31400) — x1 and x2 are x-coordinates of the left and right edges of the rectangle, and y1 and y2are y-coordinates of the bottom and top edges of the rectangle.

No two rectangles overlap (that is, there are no points that belong to the interior of more than one rectangle).

Output

In a single line print "YES", if the given rectangles form a square, or "NO" otherwise.

Sample test(s)

input

50 0 2 30 3 3 52 0 5 23 2 5 52 2 3 3

output

YES

input

40 0 2 30 3 3 52 0 5 23 2 5 5

output

NO

这是一道简单的集合的题目，想不到啊，当时大师给我提示了，下来忘写了，现在补上，其实就只需要【判断三点就行了，首先是要保证这些矩形构成的图形的面积是个完全平方数，这是正方形的首要前提，2是要保证矩形的面积合起来要和外面得大图形相同，最后就是边长相等，这样就保证他是个正方形了。

下面是代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>using namespace std;const int inf=100000000;struct square{    int lx;    int ly;    int rx;    int ry;}s[10];int main(){     int n;     while(scanf("%d",&n)!=EOF){        __int64 area=0;//刚开始时写的int,貌似出问题了        int mlx=inf,mrx=-inf;        int mly=inf,mry=-inf;           for(int i=1;i<=n;i++){              scanf("%d%d%d%d",&s[i].lx,&s[i].ly,&s[i].rx,&s[i].ry);                 mlx=min(mlx,s[i].lx);                 mrx=max(mrx,s[i].rx);                 mly=min(mly,s[i].ly);                 mry=max(mry,s[i].ry);                 area+=(s[i].rx-s[i].lx)*(s[i].ry-s[i].ly);           }           __int64 temp=sqrt(area);           if(temp*temp!=area){//如果边长不是个平方数，肯定是 错的                    cout<<"NO"<<endl;            }            else{                 if((mrx-mlx)*(mry-mly)==area&&(mrx-mlx)==(mry-mly))                      cout<<"YES"<<endl;                  else                      cout<<"NO"<<endl;            }     }     return 0;}