hdu-6341-搜索

Problem J. Let Sudoku Rotate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 530    Accepted Submission(s): 146

Problem Description

Sudoku is a logic-based, combinatorial number-placement puzzle, which is popular around the world.
In this problem, let us focus on puzzles with 16×16 grids, which consist of 4×4 regions. The objective is to fill the whole grid with hexadecimal digits, i.e. 0123456789ABCDEF, so that each column, each row, and each region contains all hexadecimal digits. The figure below shows a solved sudoku.



Yesterday, Kazari solved a sudoku and left it on the desk. However, Minato played a joke with her - he performed the following operation several times.
* Choose a region and rotate it by 90 degrees counterclockwise.
She burst into tears as soon as she found the sudoku was broken because of rotations.
Could you let her know how many operations her brother performed at least?

 

Input

The first line of the input contains an integer T (1≤T≤103) denoting the number of test cases.
Each test case consists of exactly 16 lines with 16 characters each, describing a broken sudoku.

 

Output

For each test case, print a non-negative integer indicating the minimum possible number of operations.

 

Sample Input

1 681D5A0C9FDBB2F7 0A734B62E167D9E5 5C9B73EF3C208410 F24ED18948A5CA63 39FAED5616400B74 D120C4B7CA3DEF38 7EC829A085BE6D51 B56438F129F79C2A 5C7FBC4E3D08719F AE8B1673BF42A58D 60D3AF25619C30BE 294190D8EA57264C C7D1B35606835EAB AF52A1E019BE4306 8B36DC78D425F7C9 E409492FC7FA18D2

 

Sample Output

5

Hint

The original sudoku is same as the example in the statement.

　　当时总感觉复杂度会炸，后来看题解才悟到其实由于数独的限制性较强，可以减去很大一部分无用解，然后就是普通的深搜了，

关键在于如何处理这个棋盘，我是一个一个区域处理的，枚举每个区域的旋转次数，如果和之前的不冲突就往下一个区域dfs，注意的是

函数返回的时候要将rotate过得区域再ratate回去。

　　

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
int ans,h[20],c[20],e[20][20],g[20][20];
bool vis[20];
void rotate(int a,int b){
    int i1,i2,j1,j2;
    for(i1=4*a-3,j2=4*b-3;i1<=4*a;++i1,j2++){
        for(j1=4*b-3,i2=4*a;j1<=4*b;++j1,i2--){
            g[i1][j1]=e[i2][j2];
        }
    }
    for(int i=4*a-3;i<=4*a;++i){
        for(int j=4*b-3;j<=4*b;++j){
            e[i][j]=g[i][j];
        }
    }
}
bool ok(int x,int y){
    for(int i=4*x-3;i<=4*x;++i){
        memset(vis,0,sizeof(vis));
        for(int j=1;j<=y*4;++j){
            if(vis[e[i][j]]) return 0;
            vis[e[i][j]]=1;
        }
    }
    for(int j=4*y-3;j<=4*y;++j){
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=4*x;++i){
            if(vis[e[i][j]]) return 0;
            vis[e[i][j]]=1;
        }
    }
    return 1;
}
void dfs(int x,int y,int tmp){
    if(x>4){
        if(tmp<ans) ans=tmp;
        return;
    }
    if(tmp>=ans) return;
    for(int i=0;i<4;++i){
            if(i) rotate(x,y);
            if(ok(x,y)){
                if(y==4) dfs(x+1,1,tmp+i);
                else dfs(x,y+1,tmp+i);
            }
    }
    rotate(x,y);//再转一次相当于归回原位
}
int main(){
    int t,n,m,i,j,k;
    char ch;
    cin>>t;
    while(t--){
        for(i=1;i<=16;++i){
            for(j=1;j<=16;++j){
                cin>>ch;
                e[i][j]=isdigit(ch)?ch-'0':ch-'A'+10;
            }
        }
        ans=inf;
        dfs(1,1,0);
        cout<<ans<<endl;
    }
    return 0;
}