HDOJ 5547 Sudoku (DFS 数独填数+回溯)

Sudoku

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 179    Accepted Submission(s): 67

Problem Description

Yi Sima was one of the best counselors of Cao Cao. He likes to play a funny game himself. It looks like the modern Sudoku, but smaller.

Actually, Yi Sima was playing it different. First of all, he tried to generate a 4×4 board with every row contains 1 to 4, every column contains 1 to 4. Also he made sure that if we cut the board into four2×2 pieces, every piece contains 1 to 4.

Then, he removed several numbers from the board and gave it to another guy to recover it. As other counselors are not as smart as Yi Sima, Yi Sima always made sure that the board only has one way to recover.

Actually, you are seeing this because you've passed through to the Three-Kingdom Age. You can recover the board to make Yi Sima happy and be promoted. Go and do it!!!

 

Input

The first line of the input gives the number of test cases,T(1≤T≤100).T test cases follow. Each test case starts with an empty line followed by 4 lines. Each line consist of 4 characters. Each character represents the number in the corresponding cell (one of '1', '2', '3', '4'). '*' represents that number was removed by Yi Sima.

It's guaranteed that there will be exactly one way to recover the board.

 

Output

For each test case, output one line containing Case #x:, where x is the test case number (starting from 1). Then output 4 lines with 4 characters each. indicate the recovered board.

 

Sample Input

3****234141233214*243*312*421*134*41***3*2*414*2*

 

Sample Output

Case #1:1432234141233214Case #2:1243431234212134Case #3:3412123423414123

 



题意：给一个4*4的矩阵填数，  也就是我们常说的数独

思路:此题的一个坑点就是这个数独和我们生活中玩的数独不一样，斜对角线上的数可以相同，
但是各个角落的4个数字不能相同，其他的纯暴力就好了

ac代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define MAXN 10#define MOD 1000000007#define LL long long#define INF 0xfffffffusing namespace std;char map[MAXN][MAXN];int check(int a,int b,char c){int i;for(i=0;i<4;i++){if(map[a][i]==c||map[i][b]==c)return 1;    }    if(a/2==0&&b/2==0)    {    if(map[0][0]==c||map[0][1]==c||map[1][0]==c||map[1][1]==c)    return 1;}else if(a/2==0&&b/2){if(map[0][2]==c||map[0][3]==c||map[1][2]==c||map[1][3]==c)return 1;}else if(a/2&&b/2==0){if(map[2][0]==c||map[2][1]==c||map[3][0]==c||map[3][1]==c)    return 1;}else if(a/2&&b/2){if(map[2][2]==c||map[2][3]==c||map[3][2]==c||map[3][3]==c)return 1;}return 0;}void dfs(int x,int y){int i,j;if(x==4){for(i=0;i<4;i++){for(j=0;j<4;j++)printf("%c",map[i][j]);printf("\n");}return;}if(map[x][y]!='*'){if(y==3)dfs(x+1,0);else    dfs(x,y+1);    }else{for(i=1;i<=4;i++){if(check(x,y,i+'0')==0){map[x][y]=i+'0';if(y==3)dfs(x+1,0);elsedfs(x,y+1);map[x][y]='*';//回溯}}}}int main(){int t,i;int cas=0;scanf("%d",&t);while(t--){for(i=0;i<4;i++)scanf("%s",map[i]);printf("Case #%d:\n",++cas);dfs(0,0);}    return 0;}