UVA 12291 Polyomino Composer

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 ( 1mn10)
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 

100

The Seventh Hunan Collegiate Programming Contest 
Problemsetter: Rujia Liu, Special Thanks: Yiming Li & Jane Alam Jan

這個題目的意思就是給你兩圖形，問你用兩張第二種圖能否組成第一張第一種張圖，方向不能改變。

看到這個題，我首先想到的就是枚舉(n-height)*(n-width)個點，因為m和n都很小，就枚舉法了，由於我水平很水，調了好久好久。

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int n,m;bool cover(char a[12][12],int px,int py,char b[12][12],int sx,int sy){    //功能:把子圖b從(sx,sy)到(m,m)中的‘*’貼到a圖中。    for(int i=sx,j=px;i<m;i++,j++)    {        for(int k=sy,l=py;k<m;k++,l++)        {            //如果能賦值就賦值            if(b[i][k]=='*'&&a[j][l]=='.')                a[j][l]=b[i][k];            else if(b[i][k]=='*'&&a[j][l]=='*')                return false;            //如果不能賦值則返回cover失敗。        }    }    return true;}void init(char a[12][12]){//把圖初始化為全是‘.’的圖。    for(int i=0; i<n; i++)    {        for(int j=0; j<n; j++)        {            a[i][j]='.';        }        a[i][n]='\0';    }}void copy(char a[12][12],char b[12][12]){//如命名    for(int i=0; i<n; i++)        strcpy(a[i],b[i]);}bool issame(const char a[12][12],const char b[12][12]){//判斷兩圖是否相同。    for(int i=0; i<n; i++)        if(strcmp(a[i],b[i])!=0)            return false;    return true;}int main(){    //freopen("in.txt","r",stdin);    char big[12][12],part[12][12];    char blank[12][12],tmp[12][12];    int minx,miny,maxx,maxy;    bool flag;    while(scanf("%d%d",&n,&m),m||n)    {        minx=m,miny=m,flag=0;        maxx=0,maxy=0;        for(int i=0; i<n; i++)            scanf("%s",big[i]);        for(int j=0; j<m; j++)            scanf("%s",part[j]);        for(int i=0; i<m; i++)            for(int j=0; j<m; j++)            {                if(part[i][j]=='*')                {                    if(minx>i)minx=i;if(miny>j)miny=j;                    if(maxx<i)maxx=i;if(maxy<j)maxy=j;                }            }        int height=maxx-minx,width=maxy-miny;        for(int i=0; i<n-height; i++)        {            for(int j=0; j<n-width; j++)            {                init(blank);                cover(blank,i,j,part,minx,miny);                for(int l=0; l<n-height; l++)                {                    for(int k=0; k<n-width; k++)                    {                        copy(tmp,blank);                        if(tmp[l][k]=='.'&&cover(tmp,l,k,part,minx,miny))                        {                            if(issame(big,tmp))                            {                                flag=1;                                i=100,j=100,l=100,k=100;                                //這裡給i,j,l,k賦值是用來退出多重迴圈用的。                            }                        }                    }                }            }        }        printf("%d\n",flag);    }    return 0;}