Polyomino Composer(UVA12291)

时间：2016-07-10 08:40:58 阅读：292 评论：0 收藏：0 [点我收藏+]

标签：

Description

Polyomino Composer

A polyomino is a plane geometric figure formed by joining one or more equal squares edge to edge.
- Wikipedia

Given a large polyomino and a small polyomino, your task is to determine whether you can compose the large one with two copies of the small one. The polyominoes can be translated, but not flipped or rotated. The two pieces should not overlap. The leftmost picture below is a correct way of composing the large polyomino, but the right two pictures are not. In the middle picture, one of the pieces was rotated. In the rightmost picture, both pieces are exactly identical, but they‘re both rotated from the original piece (shown in the lower-right part of the picture).

Input

There will be at most 20 test cases. Each test case begins with two integers n and m ( 1

10) in a single line. The next n lines describe the large polyomino. Each of these lines contains exactly n characters in `*‘,`.‘. A `*‘ indicates an existing square, and a `.‘ indicates an empty square. The next m lines describe the small polyomino, in the same format. These characters are guaranteed to form valid polyominoes (note that a polyomino contains at least one existing square). The input terminates with n = m = 0, which should not be processed.

Output

For each case, print `1‘ if the corresponding composing is possible, print `0‘ otherwise.

Sample Input

4 3
.**.
****
.**.
....
**.
.**
...
3 3
***
*.*
***
*..
*..
**.
4 2
****
....
....
....
*.
*.
0 0

Sample Output

1
0
0

思路：暴力枚举下即可；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<queue>
  5 #include<math.h>
  6 #include<stdlib.h>
  7 #include<string.h>
  8 char ans[20][20];
  9 char bns[20][20];
 10 char ask[20][20];
 11 char ck[20][20];
 12 bool flag=0;
 13 bool tie(int n,int m,int xx,int yy,int mxx,int myy)//先贴第一个
 14 {
 15     int i,j;
 16     for(i=n; i<=xx; i++)
 17     {
 18         for(j=m; j<=yy; j++)
 19         {
 20             int p=i-n;
 21             int q=j-m;
 22             ck[i][j]=bns[mxx+p][myy+q];
 23         }
 24     }
 25 }
 26 bool check(int n,int m,int xx,int yy,int mxx,int myy,int t)//贴第二个并判断与要求的图形是否相同
 27 {   int i,j;
 28     for(i=n; i<=xx; i++)
 29     {
 30         for(j=n; j<=yy; j++)
 31         {
 32             int p=i-n;
 33             int q=j-m;
 34             if(ask[i][j]==‘.‘&&bns[mxx+p][myy+q]==‘*‘)
 35             {
 36                 ask[i][j]=‘*‘;
 37             }
 38             else if(ask[i][j]==‘*‘&&bns[mxx+p][myy+q]==‘*‘)
 39             {
 40                 return false;
 41             }
 42         }
 43     }
 44     for(i=0;i<t;i++)
 45     {
 46         for(j=0;j<t;j++)
 47         {
 48             if(ask[i][j]!=ans[i][j])
 49                 return false;
 50         }
 51     }
 52     return true;
 53 }
 54 
 55 int main(void)
 56 {
 57     int i,j,k;
 58     int n,m;
 59     while(scanf("%d %d",&n,&m),n!=0&&m!=0)
 60     {
 61         flag=0;
 62         memset(ask,0,sizeof(ask));
 63         for(i=0; i<n; i++)
 64         {
 65             scanf("%s",ans[i]);
 66         }
 67         for(i=0; i<m; i++)
 68         {
 69             scanf("%s",bns[i]);
 70         }
 71         int x,y;
 72         int x1=0;
 73         int x2=m;
 74         int y1=0;
 75         int y2=m;
 76         for(i=0; i<m; i++)
 77         {
 78             for(j=0; j<m; j++)
 79             {
 80                 if(bns[i][j]==‘*‘)
 81                 {
 82                     if(i>x1)
 83                         x1=i;
 84                     if(j>y1)
 85                         y1=j;
 86                     if(i<x2)
 87                         x2=i;
 88                     if(j<y2)
 89                         y2=j;
 90                 }
 91             }
 92         }
 93         int xx2=x1-x2;
 94         int yy2=y1-y2;
 95         for(i=0; i<n; i++)
 96         {
 97             for(j=0; j<n; j++)
 98             {
 99                 for(int s=0; s<n; s++)
100                 {
101                     for(int uu=0; uu<n; uu++)
102                     {
103                         ck[s][uu]=‘.‘;
104                     }
105                 }
106                 if(i+xx2>n-1||j+yy2>n-1)
107                     continue;
108                 else
109                 {
110                     tie(i,j,i+xx2,j+yy2,x2,y2);
111                     for(x=0; x<n; x++)
112                     {
113                         for(y=0; y<n; y++)
114                         {
115                             for(int s=0; s<n; s++)
116                             {
117                                 for(int uu=0; uu<n; uu++)
118                                 {
119                                     ask[s][uu]=ck[s][uu];
120                                 }
121                             }
122                             if(x+xx2>n-1||y+yy2>n-1)
123                                 continue;
124                             else
125                             {
126                                 flag=check(x,y,x+xx2,y+yy2,x2,y2,n);
127                                 if(flag)
128                                 {
129                                     break;
130                                 }
131                             }
132                         }if(flag)break;
133                     }
134                 }
135                 if(flag)break;
136             }
137             if(flag)break;
138         }
139         if(flag)printf("1\n");
140         else printf("0\n");
141     }
142     return 0;
143 }

Polyomino Composer(UVA12291)

标签：

原文地址：http://www.cnblogs.com/zzuli2sjy/p/5657077.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行