HDOJ 5128 The E-pang Palace 暴力枚舉+計算幾何
來源：程序員人生 發(fā)布時間：2014-12-09 08:34:31 閱讀次數(shù)：3278次
枚舉出每個矩形,判斷相交
如果是回字型則輸出大的矩形的面積..........
The E-pang Palace
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 73 Accepted Submission(s): 26
Problem 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 magnificent 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 fight 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 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.
Input
There are no more than 15 test case.
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.
Output
For each test case, print the maximum total area of land Zhang Liang and Xiao He could get. If it was impossible for them to build two qualified fences, print "imp".
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
Source
2014ACM/ICPC亞洲區(qū)廣州站-重現(xiàn)賽（感謝華工和北京大學(xué)）
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <set>
using namespace std;
const int maxn = 31;
int n,x[maxn],y[maxn];
bool vis[201][201];
struct Point
{
int x,y;
};
struct MATRIX
{
Point p[4];
int area;
};
vector<MATRIX> mat;
bool inside(int& kind,Point p,MATRIX m)
{
int sx=m.p[0].x,dx=m.p[2].x;
int sy=m.p[0].y,dy=m.p[2].y;
if( (p.x>sx&&p.x<dx) && (p.y>sy&&p.y<dy) ) kind=1;
if( (p.x>=sx&&p.x<=dx) && (p.y>=sy&&p.y<=dy) )
return true;
return false;
}
int check(int a,int b)
{
MATRIX A=mat[a],B=mat[b];
int k0=0,k1=0,k2=0,k3=0;
bool ok0=inside(k0,A.p[0],B);
bool ok1=inside(k1,A.p[1],B);
bool ok2=inside(k2,A.p[2],B);
bool ok3=inside(k3,A.p[3],B);
/// maybe A all in side B
if(ok0||ok1||ok2||ok3)
{
if(k0==1&&k1==1&&k2==1&&k3==1)
return B.area;
else return ⑼99;
}
k0=0,k1=0,k2=0,k3=0;
ok0=inside(k0,B.p[0],A);
ok1=inside(k1,B.p[1],A);
ok2=inside(k2,B.p[2],A);
ok3=inside(k3,B.p[3],A);
/// maybe B all in side A
if(ok0||ok1||ok2||ok3)
{
if(k0==1&&k1==1&&k2==1&&k3==1)
return A.area;
else return ⑼99;
}
/// A out of B and B out of A
if(ok0==false&&ok1==false&&ok2==false&&ok3==false)
{
return A.area+B.area;
}
return ⑼99;
}
int main()
{
while(scanf("%d",&n)!=EOF&&n)
{
///init
memset(vis,0,sizeof(vis));
mat.clear();
for(int i=0;i<n;i++)
{
cin>>x[i]>>y[i];
vis[x[i]][y[i]]=true;
}
/// find all SQR
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
if(x[i]==x[j]) continue;
if(y[i]==y[j]) continue;
int sx=min(x[i],x[j]),dx=max(x[i],x[j]);
int sy=min(y[i],y[j]),dy=max(y[i],y[j]);
if(vis[sx][sy]&&vis[dx][sy]&&vis[sx][dy]&&vis[dx][dy])
{
/// this is a sqrt
MATRIX m;
m.p[0].x=sx,m.p[0].y=sy;
m.p[1].x=dx,m.p[1].y=sy;
m.p[2].x=dx,m.p[2].y=dy;
m.p[3].x=sx,m.p[3].y=dy;
m.area=(dx-sx)*(dy-sy);
mat.push_back(m);
}
}
}
int ans=⑴;
int sz=mat.size();
for(int i=0;i<sz;i++)
{
for(int j=i+1;j<sz;j++)
{
/// 判斷 4 矩形相交
ans=max(ans,check(i,j));
}
}
if(ans<0) puts("imp");
else printf("%d
",ans);
}
return 0;
}
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