UV 12113 Overlapping Squares, 12113 overlapping

UV 12113 Overlapping Squares, 12113 overlapping

Question:

There are 6 2*2 squares in total to determine whether the given shape can be formed.

Ideas:

A square has a total of 9 placement methods, and a total of 2 9 Power Placement methods for the whole map. Then, the map is represented by an array of 9*5, and the position of the square is represented by the subscript of its eight sides and four hollow subscript.

Code:

# Include <iostream>
# Include <cstring>
# Include <cstdio>
# Include <algorithm>
Using namespace std;
Int bian [9] [8] = {
},
},
},
},
},
{, 32 },
{19,21, 27,31, 36,37, 39,40 },
{, 23, 39 },
{, 35, 41,}; // edge coordinates of 9 placement methods of 2*2 square
Int nei [9] [4] = {
10, 11, 12, 20,
12,13, 14,22,
14,15, 16,24,
19,20, 21,29,
, 22,
, 33,
28, 29, 30, 38,
30, 31, 32, 40,
32, 33, 34, 42}; // null coordinates of 9 placement methods of 2*2 square

Char str [100];
Int map [45], cot [10], map1 [45];
Int read () // read data
{
Int h = 0, cnt = 0, kk = 0;
For (int I = 0; I <5; I ++)
{
If (gets (str) = NULL)
Return-1;
If (str [0] = '0 ')
Return-1;
For (int j = 0; j <9; j ++)
{
Map [kk ++] = str [j] = 32? ;
If (str [j]! = 32)
Cnt ++;
}
}
Return cnt;
}
Int count (int s) // calculates the total number of Squares used
{
Return s = 0? 0: count (s/2) + (s & 1 );
}
Void stick (int p, int & cnt) // construct the map after the p-th square is pasted.
{
Int I, j;
For (I = 0; I <8; I ++)
{
If (map1 [bian [p] [I] = 0) // if there is no edge at the side of the p-th Square, the number of edges + 1
Cnt ++;
Map1 [bian [p] [I] = 1;
}
For (I = 0; I <4; I ++)
{
If (map1 [nei [p] [I] = 1) // Number of edges in the hollow position of the p-th square if an edge exists-1

            cnt--;
        map1[nei[p][i]]=0;
    }
}
int main()
{
    int flag,cnt;
    int i,j,k;
    int h=1;
    while((flag=read())&&flag!=-1)
    {
        //cout<<flag<<endl;
        int flag1=0;
        for(i=0;i<(1<<9);i++)
        {
            int n=count(i);
            if(n>6||8*n<flag)
                continue;
            k=0;
            for(j=0;j<9;j++)
                if((i>>j)&1)
                    cot[k++]=j;
            //cout<<k<<endl;
            do
            {
                //cout<<1<<endl;
                memset(map1,0,sizeof(map1));
                cnt=0;
                //cout<<n<<endl;
                for(j=0;j<n;j++)
                {
                    int p=cot[j];//cout<<1<<endl;
                    stick(p,cnt);
                }

                if(cnt==flag)
                {
                    int flag2=1;
                    for(int l=0;l<45;l++)
                    {
                        if(map[l]!=map1[l])
                        {
                            flag2=0;break;

                        }

                    }if(flag2)
                            flag1=1;

                }
                if(flag1)
                    break;
            }
            while(next_permutation(cot,cot+k));
        }
        printf("Case %d: %s\n",h++, flag1? "Yes" : "No");
    }
    return 0;
}