A

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 ≤ m ≤ n ≤ 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

题意：在大图形中是否能刚好有两个小图形组成，没有多余，小图形只能平移，不能转动。

思路：暴力把每一个点都找一下，看是否刚好是两个小图形组成的。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<stack>
#define maxn 505
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
string s[maxn],t[maxn];
int x[maxn],y[maxn];
int l,n,m;
bool judge1(int a,int b){
    if(a<0||a>=n||b<0||b>=n||s[a][b]!='*')
        return false;
    else
        return true;
}
bool judge(int a,int b){
    for(int i=0;i<l;i++){
        if(!judge1(a+x[i],b+y[i]))return false;
    }
     for(int i=0;i<l;i++){
        s[a+x[i]][b+y[i]]='.';
    }
    return true;
}
int main(){
   while(~scanf("%d%d",&n,&m)&&n|m){
    memset(x,0,sizeof(x));
    memset(y,0,sizeof(y));
    for(int i=0;i<n;i++)cin>>s[i];
    for(int i=0;i<m;i++)cin>>t[i];
    int fx=-1,fy=-1;l=0;
    for(int i=0;i<m;i++){
        for(int j=0;j<m;j++){
            if(t[i][j]=='*'){
                if(fx==-1&&fy==-1){fx=i;fy=j;}
                else{x[l]=(i-fx);y[l]=(j-fy);l++;}
                }
            }
        }
//    for(int i=0;i<l;i++){
//        cout<<x[i]<<" "<<y[i]<<endl;
//    }
    int flag=0,ans=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(s[i][j]=='*'){
                if(judge(i,j))ans++;//cout<<ans<<endl;
                if(ans==2){flag=1;cout<<1<<endl;break;}
            }
        }if(flag==1)break;
    }
    if(flag==0)cout<<0<<endl;
   }
}