cf 194 div2 BEight Point Sets

B. Eight Point Setstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGerald is very particular to eight point sets. He thinks that any decent eight point set must consist of all pairwise intersections of three distinct integer vertical straight lines and three distinct integer horizontal straight lines, except for the averageof these nine points. In other words, there must be three integers x1, x2, x3 andthree more integers y1, y2, y3,such that x1 < x2 < x3, y1 < y2 < y3 andthe eight point set consists of all points (xi, yj) (1 ≤ i, j ≤ 3),except for point (x2, y2).You have a set of eight points. Find out if Gerald can use this set?InputThe input consists of eight lines, the i-th line contains two space-separated integers xi and yi (0 ≤ xi, yi ≤ 106).You do not have any other conditions for these points.OutputIn a single line print word "respectable", if the given set of points corresponds to Gerald's decency rules, and "ugly"otherwise.Sample test(s)input

0 0
0 1
0 2
1 0
1 2
2 0
2 1
2 2

output

respectable

input

0 0
1 0
2 0
3 0
4 0
5 0
6 0
7 0

output

ugly

input

1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2

output

ugly

表示看了半天题，都没看懂啥意思，最后靠着小胖翻译的帮助才将这个题看懂，哎，酱油啊，下面是代码:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;struct node{int x;int y;}p[10];bool cmp(node a,node b){if(a.x!=b.x)return a.x<b.x;return a.y<b.y;}int main(){for(int i=0;i<8;i++)scanf("%d%d",&p[i].x,&p[i].y);sort(p,p+8,cmp);bool ans=true;if(p[0].x != p[1].x || p[1].x != p[2].x)ans = false ;if(p[3].x != p[4].x)ans = false ;if(p[5].x != p[6].x || p[6].x != p[7].x)ans = false ;int y1 = p[0].y ;int y2 = p[1].y ;int y3 = p[2].y ;if(y1 >= y2 || y2 >= y3)ans = false ;if(p[3].y != y1 || p[4].y != y3)ans = false ;if(p[5].y != y1 || p[6].y != y2 || p[7].y != y3)ans = false ;if(ans)cout<<"respectable"<<endl;elsecout<<"ugly"<<endl;return 0 ;}