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
mn10) 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 "); 140 else printf("0 "); 141 } 142 return 0; 143 }