题目链接：http://acm.hdu.edu.cn/showproblem.php?
题面：
The E-pang Palace
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 1646 Accepted Submission(s): 1226
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 fire lasted for more than three months, renouncing the end of Qin dynasty.
Several years later, Liu Bang defeated Xiangyu and became the first 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 figures below shows 3 situations which are not qualified(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.
For each test case:
The first 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.
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
2 imp
给定一些点。要以这些点为四个顶点构造矩形，求构造两个矩形不相交（相交关系见图示）。且面积最大。
不可能则输出imp。
比較暴力就好，有点类似上周BC的一道题的做法，直接枚举两个对角线点，然后看另外两个点是否存在，存在的话。继续和还有一个矩形推断相对位置关系。
可是要小心一个矩形包括在另外一个矩形中，也是能够的，面积仅仅算大的那个。具体看代码。
代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
using namespace std;
struct point
{
int x,y;
}store[35];
bool map[210][210];
int max(int a,int b)
{
return a>b?a:b;
}
int min(int a,int b)
{
return a<b?a:b;
}
int judge(int a,int b,int c,int d)
{
int res;
//下面分别为两个矩形左下和右上位置坐标
int lx1,lx2,ly1,ly2,rx1,rx2,ry1,ry2;
lx1=min(store[a].x,store[b].x);
ly1=min(store[a].y,store[b].y);
lx2=min(store[c].x,store[d].x);
ly2=min(store[c].y,store[d].y);
rx1=max(store[a].x,store[b].x);
ry1=max(store[a].y,store[b].y);
rx2=max(store[c].x,store[d].x);
ry2=max(store[c].y,store[d].y);
//是同一个矩形
if(lx1==lx2&&ly1==ly2&&rx1==rx2&&ry1==ry2)
return 0;
//一个全然在另外一个上面或者右边，那么就是两个矩形加
if((ry1<ly2)||(ry2<ly1)||(lx1>rx2)||(lx2>rx1))
{
res=(rx1-lx1)*(ry1-ly1)+(rx2-lx2)*(ry2-ly2);
return res;
}
//假设一个全然包括在另外一个里面
if(lx2<lx1&&ly2<ly1&&rx1<rx2&&ry1<ry2)
{
res=(rx2-lx2)*(ry2-ly2);
return res;
}
//假设一个全然包括在另外一个里面
if(lx1<lx2&&ly1<ly2&&rx2<rx1&&ry2<ry1)
{
res=(rx1-lx1)*(ry1-ly1);
return res;
}
//否则即为相交
else
return 0;
}
int main()
{
int n,a,b,c,d,e,f,g,h;
int area;
while(scanf("%d",&n)&&n)
{
area=0;
memset(map,0,sizeof(map));
//读入
for(int i=0;i<n;i++)
{
scanf("%d%d",&store[i].x,&store[i].y);
//将地图上对应的点标记为有
map[store[i].x][store[i].y]=1;
}
//没8个点，就肯定不行
if(n<8)
{
printf("imp\n");
continue;
}
else
{
for(int i=0;i<n-1;i++)
{
for(int j=i+1;j<n;j++)
{
//由于枚举的是对角线。所以在一条边上就跳过
if((store[i].x==store[j].x)||(store[i].y==store[j].y))
continue;
else
{
a=store[i].x;
b=store[i].y;
c=store[j].x;
d=store[j].y;
//假设除了对角线之外，另外两点存在
if(map[c][b]&&map[a][d])
{
for(int k=0;k<n-1;k++)
{
//不与原来两点反复
if((k!=i)&&(k!=j))
{
for(int m=k+1;m<n;m++)
{
//不与原来两点反复
if((m!=i)&&(m!=j))
{
//仅仅枚举对角线
if((store[k].x==store[m].x)||(store[k].y==store[m].y))
continue;
else
{
e=store[k].x;
f=store[k].y;
g=store[m].x;
h=store[m].y;
//假设除了对角线之外。另外两点存在,这时才推断两个矩形关系
if(map[g][f]&&map[e][h])
{
area=max(area,judge(i,j,k,m));
}
}
}
}
}
}
}
else
continue;
}
}
}
if(area==0)
printf("imp\n");
else
printf("%d\n",area);
}
}
return 0;
}
