poj 2517

 

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Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 134		Accepted: 13

Description

Given an N * N matrix A, whose elements are positive integers and not larger than 10000. There are also some '*'s in matrix A. Your program must find such three (perhaps overlapping) rectangles that they would contain all the '*'s which are in matrix A and the areas of these three rectangles (the area of one rectangle define as the number of elements it covers) must be non-negative and not exceed a given integer M. 

If several solutions exist, the program should find a solution with minimal total costs of these three rectangles. The cost of one rectangle defines as the sum of all the numbers it contains in matrix A.

Input

The first line of the input is an integer X representing the number of test cases. The following X blocks each represents a test case. 

The first line of each block contains two numbers N and M (1 <= N <= 30, 0 <= M <= N * N) representing the size of the matrix and the maximum area of each rectangle. The second line contains a number C (0 <= C <= N * N), representing the number of '*'s in matrix A. The following C lines each contains two numbers X and Y (1 <= X, Y <= N), representing A[X][Y] (A[X][Y] represents the element in the X-th row and Y-th column)contains a '*'. The following N lines each contains N numbers, representing the matrix A. 

Output

For each block output one line, which contains an integer representing the minimum total costs, or 'Impossible' (without quote) if no solution exists.

Sample Input

5
1 1
0
9
1 1
1
1 1
9
5 6
5
1 1
3 4
4 3
4 5
5 4
5 3 1 1 1
3 1 1 1 1
1 1 1 2 1
1 1 2 5 2
1 1 1 2 1
5 3
5
1 1
3 4
4 3
4 5
5 4
5 3 1 1 1
3 1 1 1 1
1 1 1 2 1
1 1 2 5 2
1 1 1 2 1
5 2
4
1 1
3 4
4 3
4 5
5 3 1 1 1
3 1 1 1 1
1 1 1 2 1
1 1 2 5 2
1 1 1 2 1

Sample Output

0
9
20
23
Impossible

Source

POJ Monthly--2005.07.31, LouTianCheng

分析：矩形分割，分情况讨论。。。具体太复杂，就不说了。。。

过下面几组数据大概就能过了，就这几种情况

5 6

7

2 2

2 4

3 1

3 3

3 5

4 2

4 4

9 9 9 9 9

9 1 9 1 9

1 1 1 1 1

9 1 9 1 9

9 9 9 9 9

5 6

7

1 3

2 2

2 4

3 3

4 2

4 4

5 3

9 9 1 9 9

9 1 1 1 9

9 9 1 9 9

9 1 1 1 9

9 9 1 9 9

5 3

5

2 2

3 1

3 3

3 5

4 2

9 9 9 9 9

9 1 9 1 9

1 1 1 1 1

9 1 9 1 9

9 9 9 9 9

5 9

6

1 3

2 4

3 1

3 5

4 2

5 3

1 1 1 9 9

1 1 1 1 9

1 1 1 1 1

9 1 1 1 1

9 9 1 1 1

5 9

6

1 3

2 2

3 1

3 5

4 4

5 3

9 9 1 1 1

9 1 1 1 1

1 1 1 1 1

1 1 1 1 9

1 1 1 9 9

#include <iostream> using namespace std; int tot[33][33]; char get[4][33][33][33][33]; int n,m,c,x,x0,x1,x2,x3,y0,y1,y2,y3,ans,now1,now2,now3; bool flag; char get1(char a,char b,char c) { if(b && ((b<a)||(a==0)))a=b; if(c && ((c<a)||(a==0)))a=c; return(a); } char get2(char a,char b,char c){return(max(a,max(b,c)));} int count(int x1,int x2,int y1,int y2) { if(x1==0)return 0; if((x2-x1+1)*(y2-y1+1)>m){flag=1;return 0;} return(tot[x2][y2]-tot[x2][y1-1]-tot[x1-1][y2]+tot[x1-1][y1-1]); } int main(int argc, char** argv) { //freopen("car.in","r",stdin); //freopen("car.out","w",stdout); scanf("%d",&x); while(x-->0) { scanf("%d%d%d",&n,&m,&c); memset(get,0,sizeof(get)); memset(tot,0,sizeof(tot)); while(c-->0) { scanf("%d%d",&x1,&y1); get[0][x1][y1][x1][y1]=x1; get[1][x1][y1][x1][y1]=x1; get[2][x1][y1][x1][y1]=y1; get[3][x1][y1][x1][y1]=y1; } for(x1=1;x1<=n;++x1) for(y1=1;y1<=n;++y1) scanf("%d",&tot[x1][y1]),tot[x1][y1]+=tot[x1][y1-1]; for(x1=2;x1<=n;++x1) for(y1=1;y1<=n;++y1) tot[x1][y1]+=tot[x1-1][y1]; for(x1=1;x1<=n;++x1) for(y1=1;y1<=n;++y1) for(x2=x1;x2<=n;++x2) for(y2=y1;y2<=n;++y2) { get[0][x1][y1][x2][y2]=get1(get[0][x1][y1][x2-1][y2],get[0][x1][y1][x2][y2-1],get[0][x2][y2][x2][y2]); get[1][x1][y1][x2][y2]=get2(get[1][x1][y1][x2-1][y2],get[1][x1][y1][x2][y2-1],get[1][x2][y2][x2][y2]); get[2][x1][y1][x2][y2]=get1(get[2][x1][y1][x2-1][y2],get[2][x1][y1][x2][y2-1],get[2][x2][y2][x2][y2]); get[3][x1][y1][x2][y2]=get2(get[3][x1][y1][x2-1][y2],get[3][x1][y1][x2][y2-1],get[3][x2][y2][x2][y2]); } x0=get[0][1][1][n][n]; y0=get[2][1][1][n][n]; x3=get[1][1][1][n][n]; y3=get[3][1][1][n][n]; ans=INT_MAX; flag=0; now1=count(x0,x3,y0,y3); if(!flag)ans=min(ans,now1); for(x1=x0;x1<=x3;++x1) for(y1=y0;y1<=y3;++y1) for(x2=x0;x2<=x3;++x2) for(y2=y0;y2<=y3;++y2) { flag=0; now1=count(get[0][x0][y0][x1][y1],get[1][x0][y0][x1][y1], get[2][x0][y0][x1][y1],get[3][x0][y0][x1][y1]); if(flag)continue; now2=count(get[0][x2][y2][x3][y3],get[1][x2][y2][x3][y3], get[2][x2][y2][x3][y3],get[3][x2][y2][x3][y3]); if(flag)continue; int xx1=0,xx2=0,yy1=0,yy2=0; if((x1<x2)&&(y2<=y1)) { xx1=get[0][x1+1][y2][x2-1][y1]; xx2=get[1][x1+1][y2][x2-1][y1]; yy1=get[2][x1+1][y2][x2-1][y1]; yy2=get[3][x1+1][y2][x2-1][y1]; } if((x2<=x1)&&(y1<y2)) { xx1=get[0][x2][y1+1][x1][y2-1]; xx2=get[1][x2][y1+1][x1][y2-1]; yy1=get[2][x2][y1+1][x1][y2-1]; yy2=get[3][x2][y1+1][x1][y2-1]; } xx1=get1(xx1,get[0][x0][y1+1][x2-1][y3],get[0][x1+1][y0][x3][y2-1]); xx2=get2(xx2,get[1][x0][y1+1][x2-1][y3],get[1][x1+1][y0][x3][y2-1]); yy1=get1(yy1,get[2][x0][y1+1][x2-1][y3],get[2][x1+1][y0][x3][y2-1]); yy2=get2(yy2,get[3][x0][y1+1][x2-1][y3],get[3][x1+1][y0][x3][y2-1]); now3=count(xx1,xx2,yy1,yy2); if(!flag)ans=min(ans,now1+now2+now3); } for(x1=x0;x1<=x3;++x1) for(y1=y0;y1<=y3;++y1) for(x2=x0;x2<=x3;++x2) for(y2=y0;y2<=y3;++y2) { flag=0; now1=count(get[0][x1][y0][x3][y1],get[1][x1][y0][x3][y1], get[2][x1][y0][x3][y1],get[3][x1][y0][x3][y1]); if(flag)continue; now2=count(get[0][x0][y2][x2][y3],get[1][x0][y2][x2][y3], get[2][x0][y2][x2][y3],get[3][x0][y2][x2][y3]); if(flag)continue; int xx1=0,xx2=0,yy1=0,yy2=0; if((x1>x2)&&(y1>=y2)) { xx1=get[0][x2+1][y2][x1-1][y1]; xx2=get[1][x2+1][y2][x1-1][y1]; yy1=get[2][x2+1][y2][x1-1][y1]; yy2=get[3][x2+1][y2][x1-1][y1]; } if((x1<=x2)&&(y1<y2)) { xx1=get[0][x1][y1+1][x2][y2-1]; xx2=get[1][x1][y1+1][x2][y2-1]; yy1=get[2][x1][y1+1][x2][y2-1]; yy2=get[3][x1][y1+1][x2][y2-1]; } xx1=get1(xx1,get[0][x0][y0][x1-1][y2-1],get[0][x2+1][y1+1][x3][y3]); xx2=get2(xx2,get[1][x0][y0][x1-1][y2-1],get[1][x2+1][y1+1][x3][y3]); yy1=get1(yy1,get[2][x0][y0][x1-1][y2-1],get[2][x2+1][y1+1][x3][y3]); yy2=get2(yy2,get[3][x0][y0][x1-1][y2-1],get[3][x2+1][y1+1][x3][y3]); now3=count(xx1,xx2,yy1,yy2); if(!flag)ans=min(ans,now1+now2+now3); } for(x1=x0;x1<=x3;++x1) for(y1=x1;y1<=x3;++y1) for(x2=y1+1;x2<=x3;++x2) for(y2=x2;y2<=x3;++y2) { flag=0; now1=count(get[0][x1][y0][y1][y3],get[1][x1][y0][y1][y3], get[2][x1][y0][y1][y3],get[3][x1][y0][y1][y3]); if(flag)continue; now2=count(get[0][x2][y0][y2][y3],get[1][x2][y0][y2][y3], get[2][x2][y0][y2][y3],get[3][x2][y0][y2][y3]); if(flag)continue; int xx1=0,xx2=0,yy1=0,yy2=0; xx1=get1(get[0][x0][y0][x1-1][y3],get[0][y1+1][y0][x2-1][y3],get[0][y2+1][y0][x3][y3]); xx2=get2(get[1][x0][y0][x1-1][y3],get[1][y1+1][y0][x2-1][y3],get[1][y2+1][y0][x3][y3]); yy1=get1(get[2][x0][y0][x1-1][y3],get[2][y1+1][y0][x2-1][y3],get[2][y2+1][y0][x3][y3]); yy2=get2(get[3][x0][y0][x1-1][y3],get[3][y1+1][y0][x2-1][y3],get[3][y2+1][y0][x3][y3]); now3=count(xx1,xx2,yy1,yy2); if(!flag)ans=min(ans,now1+now2+now3); } for(x1=y0;x1<=y3;++x1) for(y1=x1;y1<=y3;++y1) for(x2=y1+1;x2<=y3;++x2) for(y2=x2;y2<=y3;++y2) { flag=0; now1=count(get[0][x0][x1][x3][y1],get[1][x0][x1][x3][y1], get[2][x0][x1][x3][y1],get[3][x0][x1][x3][y1]); if(flag)continue; now2=count(get[0][x0][x2][x3][y2],get[1][x0][x2][x3][y2], get[2][x0][x2][x3][y2],get[3][x0][x2][x3][y2]); if(flag)continue; int xx1=0,xx2=0,yy1=0,yy2=0; xx1=get1(get[0][x0][y0][x3][x1-1],get[0][x0][y1+1][x3][x2-1],get[0][x0][y2+1][x3][y3]); xx2=get2(get[1][x0][y0][x3][x1-1],get[1][x0][y1+1][x3][x2-1],get[1][x0][y2+1][x3][y3]); yy1=get1(get[2][x0][y0][x3][x1-1],get[2][x0][y1+1][x3][x2-1],get[2][x0][y2+1][x3][y3]); yy2=get2(get[3][x0][y0][x3][x1-1],get[3][x0][y1+1][x3][x2-1],get[3][x0][y2+1][x3][y3]); now3=count(xx1,xx2,yy1,yy2); if(!flag)ans=min(ans,now1+now2+now3); } if(ans<INT_MAX)printf("%d/n",ans); else printf("Impossible/n"); } return 0; }