CDOJ 1222--Sudoku【DFS && 数独】
来源:互联网 发布:网络直播开场白 编辑:程序博客网 时间:2024/06/05 19:46
H - Sudoku
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
4×4
board with every row contains
1
1
to
4
4
, every column contains
1
1
to
4
4
.
Also he made sure that if we cut the board into four
2×2
2×2
pieces, every piece contains
1
1
to
4
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
T
(
1≤T≤100
1≤T≤100
).
T
T
test cases follow. Each test case starts with an empty line followed by
4
4
lines.
Each line consist of
4
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
x
is the test case number (starting from
1
1
).
Then output
4
4
lines with
4
4
characters each. indicate the recovered board.
Sample Input
3
*****
2341
4123
3214
*243
*312
*421
*134
*41*
**3*
2*41
4*2*
Sample Output
Case #1:
1432
2341
4123
3214
Case #2:
1243
4312
3421
2134
Case #3:
3412
1234
2341
4123
呵呵,数读,注意题中条件,四个不一样,横向纵向不同
#include<stdio.h>#define max 4int b[6][6];int v[6][6][4]={0};char xx[6][6]={0};/*int jb(int x,int y,int mun){ for(int i=0;i<4;i++) { if(b[x][i]==mun) { return 0; } if(b[i][y]==mun) { return 0; } } return 1;}*/bool jb(int x,int y,int a){ for(int i=0;i<4;i++) { if(b[x][i]==a) { return false; } if(b[i][y]==a) { return false; } } if(x>=0&&x<=1){ if(y>=0&&y<=1){ if(b[0][0]==a||b[0][1]==a||b[1][0]==a||b[1][1]==a)return false; } else if(b[0][2]==a||b[0][3]==a||b[1][2]==a||b[1][3]==a)return false; } else { if(y>=0&&y<=1){ if(b[2][0]==a||b[2][1]==a||b[3][0]==a||b[3][1]==a)return false; } else if(b[2][2]==a||b[2][3]==a||b[3][2]==a||b[3][3]==a)return false; } return true;}int flag=0;void bfs(int x,int y){ if(flag) return; if(x==4) { for(int i=0;i<max;i++) { for(int j=0;j<max;j++) { printf("%d",b[i][j]); } printf("\n"); } flag=1; return ; } if(b[x][y]!=-1) { if(y==3) { bfs(x+1,0); } else { bfs(x,y+1); } } else { for(int i=1;i<=4;i++) { if(jb(x,y,i)) { b[x][y]=i; // printf("%d",b[x][y]); if(y==3) { bfs(x+1,0); } else { bfs(x,y+1); } b[x][y]=-1; } } }}int main(){ int t; scanf("%d",&t); int tt=0; getchar(); while(t--) { tt++; for(int i=0;i<4;i++) { scanf("%s",&xx[i]); for(int j=0;j<4;j++) { if(xx[i][j]!='*') { b[i][j]=xx[i][j]-'0'; } else { b[i][j]=-1; } // printf("%d",b[i][j]); } } flag=0; printf("Case #%d:\n",tt); bfs(0,0); } return 0;}
- CDOJ 1222--Sudoku【DFS && 数独】
- CDOJ 1222--Sudoku【DFS && 数独】
- Sudoku 数独问题(dfs)
- hdu 3111 Sudoku(DFS数独)
- HDU 1426 Sudoku Killer DFS数独
- POJ 2676 Sudoku (数独 DFS)
- [DFS]poj 2676 Sudoku 数独问题
- HDU 1426 Sudoku Killer (DFS 数独)
- POJ 2676-Sudoku(DFS-数独)
- hdu1426数独游戏Sudoku Killer DFS
- 【POJ 2676】Sudoku数独(DFS)
- hdu 1426 Sudoku Killer(DFS 数独问题)
- hdu 2780 Su-Su-Sudoku(DFS数独)
- Sudoku Solver 破解数独 @LeetCode 附DFS感想
- Sudoku Killer(数独DFS+恶心的输入输出!!!)
- hdu1426 Sudoku Killer(数独游戏)--DFS
- hdu 1426 Sudoku Killer(数独 dfs)
- POJ 2676 Sudoku (数独求解器 DFS)
- notepad++的好用快捷键整理
- LDR,ADRL,ADR,NOP伪指令
- C++抽象编程——面向对象(2)——结构体与类
- Raft论文翻译
- 非常简单的为Android设置TabBar
- CDOJ 1222--Sudoku【DFS && 数独】
- Binder源码解析(从客户端到服务端代码流程)
- Notepad++ 快捷键列表及自定义快捷键
- 数据操作语句-1
- 15题 flag的应用
- "数据结构动画演示"文件分享
- 对给定值进行操作
- vitamio5.x的简单使用
- 求助