poj 2836 Rectangular Covering(状态压缩dp)

n points are given on the Cartesian plane. Now you have to use some rectangles whose sides are parallel to the axes to cover them. Every point must be covered. And a point can be covered by several rectangles. Each rectangle should cover at least two points including those that fall on its border. Rectangles should have integral dimensions. Degenerate cases (rectangles with zero area) are not allowed. How will you choose the rectangles so as to minimize the total area of them?

The input consists of several test cases. Each test cases begins with a line containing a single integer n (2 ≤ n ≤ 15). Each of the next n lines contains two integers x, y (−1,000 ≤ x, y ≤ 1,000) giving the coordinates of a point. It is assumed that no two points are the same as each other. A single zero follows the last test case.

Output the minimum total area of rectangles on a separate line for each test case.

2
0 1
1 0
0

1

The total area is calculated by adding up the areas of rectangles used.

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define inf 1<<29
#define N 16
#define M 1<<N
int x
,y
;
int state

;
int area

;
int dp[M];
int main()
{
int n;
while(scanf("%d",&n)==1){
if(n==0)
break;
memset(state,0,sizeof(state));
for(int i=0;i<n;i++){
scanf("%d%d",&x[i],&y[i]);
}
for(int i=0;i<n;i++){
for(int j=0;j<i;j++){
int lx=min(x[i],x[j]),rx=max(x[i],x[j]);
int ly=min(y[i],y[j]),ry=max(y[i],y[j]);
for(int k=0;k<n;k++){
if(lx<=x[k] && x[k]<=rx && ly<=y[k] && y[k]<=ry){
state[i][j]+=1<<k;
}
}
int a=rx-lx?rx-lx:1;
int b=ry-ly?ry-ly:1;
area[i][j]=a*b;
}
}

for(int i=0;i<(1<<n);i++)
dp[i]=inf;
dp[0]=0;
for(int i=0;i<(1<<n);i++){
for(int j=0;j<n;j++){
for(int k=0;k<n;k++){
int T= i|state[j][k];
dp[T]=min(dp[T],dp[i]+area[j][k]);
}
}
}
printf("%d\n",dp[(1<<n)-1]);
}
return 0;
}