The E-pang Palace

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=217902

Description

E-pang Palace was built in Qin dynasty by Emperor Qin Shihuang in Xianyang, Shanxi Province. It
was the largest palace ever built by human. It was so large and so magnicent that after many years
of construction, it still was not completed. Building the great wall, E-pang Palace and Qin Shihuang's
tomb cost so much labor and human lives that people rose to ght against Qin Shihuang's regime.
Xiang Yu and Liu Bang were two rebel leaders at that time. Liu Bang captured Xianyang | the
capital of Qin. Xiang Yu was very angry about this, and he commanded his army to march to Xianyang.
Xiang Yu was the bravest and the strongest warrior at that time, and his army was much more than
Liu Bang's. So Liu Bang was frighten and retreated from Xianyang, leaving all treasures in the grand
E-pang Palace untouched. When Xiang Yu took Xianyang, he burned E-pang Palce. The re lasted
for more than three months, renouncing the end of Qin dynasty.
Several years later, Liu Bang defeated Xiangyu and became the rst emperor of Han dynasty. He
went back to E-pang Palace but saw only some pillars left. Zhang Liang and Xiao He were Liu Bang's
two most important ministers, so Liu Bang wanted to give them some awards. Liu Bang told them:
\You guys can make two rectangular fences in E-pang Palace, then the land inside the fences will
belongs to you. But the corners of the rectangles must be the pillars left on the ground, and two fences
can't cross or touch each other."
To simplify the problem, E-pang Palace can be consider as a plane, and pillars can be considered as
points on the plane. The fences you make are rectangles, and you MUST make two rectangles. Please
note that the rectangles you make must be parallel to the coordinate axes.
The gures below shows 3 situations which are not qualied (Thick dots stands for pillars):
Zhang Liang and Xiao He wanted the total area of their land in E-pang Palace to be maximum.
Please bring your computer and go back to Han dynasty to help them so that you may change the
history.

Input

There are no more than 15 test case.
For each test case: The rst line is an integer N, meaning that there are N pillars left in E-pang
Palace(4 N 30). Then N lines follow. Each line contains two integers x and y (0 x; y 200),
indicating a pillar's coordinate. No two pillars has the same coordinate.
The input ends by N = 0.

Output

Sample Input

8
0 0
1 0
0 1
1 1
0 2
1 2
0 3
1 3
8
0 0
2 0
0 2
2 2
1 2
3 2
1 3
3 3
0

Sample Output

2

imp

HINT

题意

平面给你n个点，让你找到两个和坐标轴平行的矩形，然后要求这两个矩形要么内含，要么相离

然后输出能够组成的最大面积

题解：

首先我们枚举矩形，只用n^2枚举就好了

只用枚举对角线，然后枚举之后，我们就暴力去判断就好了

判断两种情况，想清楚了，还是很简单的

代码：

#include<iostream>
#include<stdio.h>
#include<iostream>
#include<cstring>
using namespace std;int mp[250][250];
struct node
{int x,y;
};
node p[34];
int In(node a,node c,node b)
{if(a.x<=b.x&&a.x>=c.x&&a.y<=b.y&&a.y>=c.y)return 1;return 0;
}
int No_In(node a,node c,node b)
{if(a.x<b.x&&a.x>c.x&&a.y<b.y&&a.y>c.y)return 1;return 0;
}
int check(node a,node b,node c,node d)
{node t1[4],t2[4];t1[0].x=min(a.x,b.x);t1[0].y=min(a.y,b.y);t1[1].x=max(a.x,b.x);t1[1].y=min(a.y,b.y);t1[2].x=min(a.x,b.x);t1[2].y=max(a.y,b.y);t1[3].x=max(a.x,b.x);t1[3].y=max(a.y,b.y);t2[0].x=min(c.x,d.x);t2[0].y=min(c.y,d.y);t2[1].x=max(c.x,d.x);t2[1].y=min(c.y,d.y);t2[2].x=min(c.x,d.x);t2[2].y=max(c.y,d.y);t2[3].x=max(c.x,d.x);t2[3].y=max(c.y,d.y);for(int i=0;i<4;i++){//cout<<t1[i].x<<" "<<t1[i].y<<endl;if(mp[t1[i].x][t1[i].y]==0)return 0;}for(int i=0;i<4;i++){//cout<<t2[i].x<<" "<<t2[i].y<<endl;if(mp[t2[i].x][t2[i].y]==0)return 0;}if(t1[0].x==t1[1].x)return 0;if(t1[0].y==t1[2].y)return 0;if(t2[0].x==t2[1].x)return 0;if(t2[0].y==t2[2].y)return 0;int flag = 0;for(int i=0;i<4;i++)if(No_In(t1[i],t2[0],t2[3]))flag++;if(flag==4)return (t2[3].x-t2[0].x)*(t2[3].y-t2[0].y);flag = 0;for(int i=0;i<4;i++)if(No_In(t2[i],t1[0],t1[3]))flag++;if(flag==4)return (t1[3].x-t1[0].x)*(t1[3].y-t1[0].y);for(int i=0;i<4;i++)if(In(t1[i],t2[0],t2[3]))return 0;for(int i=0;i<4;i++)if(In(t2[i],t1[0],t1[3]))return 0;return (t2[3].x-t2[0].x)*(t2[3].y-t2[0].y) + (t1[3].x-t1[0].x)*(t1[3].y-t1[0].y);
}
int main()
{int n;while(scanf("%d",&n)!=EOF){memset(mp,0,sizeof(mp));if(n==0)break;for(int i=1;i<=n;i++){scanf("%d%d",&p[i].x,&p[i].y);mp[p[i].x][p[i].y]=1;}int ans = 0;for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){for(int t=1;t<=n;t++){for(int k=t+1;k<=n;k++){if(i==t&&j==k)continue;int Area = check(p[i],p[j],p[t],p[k]);if(Area!=0){//printf("%d %d %d %d %d %d %d %d\n",p[i].x,p[i].y,p[j].x,p[j].y,p[t].x,p[t].y,p[k].x,p[k].y);//cout<<Area<<endl;
                        }ans = max(ans,Area);}}}}if(ans==0)printf("imp\n");else printf("%d\n",ans);}
}