H - Sudoku

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 four 2×2 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

****
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 题意:给你4*4的图,数独游戏,4个2*2的块独立,每行每列独立
题解:暴力
///
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a))
#define TS printf("111111\n")
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define inf 100000
inline ll read()
{
ll x=,f=;
char ch=getchar();
while(ch<''||ch>'')
{
if(ch=='-')f=-;
ch=getchar();
}
while(ch>=''&&ch<='')
{
x=x*+ch-'';
ch=getchar();
}
return x*f;
}
//****************************************
#define maxn 5
bool flag;
int ans=;
char mp[maxn][maxn];
bool test()
{
if((mp[][]-'')*(mp[][]-'')*(mp[][]-'')*(mp[][]-'')!=)return ;
//if((mp[0][2]+mp[0][3]+mp[1][2]+mp[1][3]-'0'-'0'-'0'-'0')!=10)return 0;
if((mp[][]-'')*(mp[][]-'')*(mp[][]-'')*(mp[][]-'')!=)return ;
// if((mp[2][0]+mp[2][1]+mp[3][0]+mp[3][1]-'0'-'0'-'0'-'0')!=10)return 0; if((mp[][]-'')*(mp[][]-'')*(mp[][]-'')*(mp[][]-'')!=)return ;
// if((mp[2][2]+mp[2][3]+mp[3][2]+mp[3][3]-'0'-'0'-'0'-'0')!=10)return 0; if((mp[][]-'')*(mp[][]-'')*(mp[][]-'')*(mp[][]-'')!=)return ;
return ;
}
void dfs(int x,int y,int t){
if(t==ans){
// cout<<1<<endl;
if(test()){
for(int i=;i<;i++){
for(int j=;j<;j++){
printf("%c",mp[i][j]);
}
cout<<endl;
flag=;
}
}
return ;
} if(flag)return ; if(mp[x][y] == '*'){
for(int k=;k<=;k++){
int flags=;
for(int i=;i<;i++){if(y!=i&&mp[x][i]==k+'')flags=;}
for(int i=;i<;i++){if(x!=i&&mp[i][y]==k+'')flags=;}
if(flags)continue; mp[x][y]=k+'';
if(y==){dfs(x+,,t+);}
else {dfs(x,y+,t+);}
if(flag)return ;
mp[x][y]='*';
}
}
else {
if(y==)
dfs(x+,,t);
else dfs(x,y+,t);
} }
int main(){
int T=read();
int oo=;
while(T--){
ans=;flag=;
for(int i=;i<;i++){
scanf("%s",mp[i]);
for(int j=;j<;j++){
if(mp[i][j]=='*')ans++;
}
}
printf("Case #%d:\n",oo++);
dfs(,,);
}
return ;
}

代码

2015南阳CCPC H - Sudoku 数独的更多相关文章

  1. 2015南阳CCPC H - Sudoku 暴力

    H - Sudoku Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description Yi Sima was one of the best cou ...

  2. 2015南阳CCPC E - Ba Gua Zhen 高斯消元 xor最大

    Ba Gua Zhen Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description During the Three-Kingdom perio ...

  3. 2015南阳CCPC F - The Battle of Guandu 多源多汇最短路

    The Battle of Guandu Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description In the year of 200, t ...

  4. 2015南阳CCPC L - Huatuo's Medicine 水题

    L - Huatuo's Medicine Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description Huatuo was a famous ...

  5. 2015南阳CCPC G - Ancient Go 暴力

    G - Ancient Go Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description Yu Zhou likes to play Go wi ...

  6. 2015南阳CCPC D - Pick The Sticks dp

    D - Pick The Sticks Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description The story happened lon ...

  7. 2015南阳CCPC A - Secrete Master Plan 水题

    D. Duff in Beach Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description Master Mind KongMing gave ...

  8. 2015南阳CCPC G - Ancient Go dfs

    G - Ancient Go Description Yu Zhou likes to play Go with Su Lu. From the historical research, we fou ...

  9. 2015南阳CCPC D - Pick The Sticks 背包DP.

    D - Pick The Sticks Description The story happened long long ago. One day, Cao Cao made a special or ...

随机推荐

  1. POJ_2239_Selecting Courses

    题意:一周上7天课,每天12节课,学校最多开设300节不同的课,每周每种课可以只有一个上课时间或者多个上课时间(上课内容一样),问一周最多可以选多少节课. 分析:二分图最大匹配,将一周84个时间点和可 ...

  2. Scrapy框架 之某网站产品采集案例

    一.创建项目 第一步:scrapy startproject boyuan 第二步:cd boyuan scrapy genspider product -t crawl  boyuan.com 如图 ...

  3. 413 Request Entity Too Large报错处理

    修改nginx配置   这是最简单的一个做法,着报错原因是nginx不允许上传配置过大的文件,那么件把nginx的上传大小配置调高就好.    1.打开nginx主配置文件nginx.conf,一般在 ...

  4. CAD利用Select2得到所有实体(网页版)

    主要用到函数说明: IMxDrawSelectionSet::Select2 构造选择集.详细说明如下: 参数 说明 [in] MCAD_McSelect Mode 构造选择集方式 [in] VARI ...

  5. Python自学-1-基本概念问题

    C语言适合开发那些追求运行速度.充分发挥硬件性能的程序. Python是用来编写应用程序的高级编程语言. Python提供了 第三方库 & 基础代码库(覆盖了网络.文件.GUI.数据库.文本等 ...

  6. rem2

    html{font-size:50px;}body{font-size:24px;}@media screen and (min-width:320px){ html{font-size:21.333 ...

  7. IP地址、MAC地址、ARP地址解析协议

    互联网中一台主机要和另一台主机实现通信首先需要知道彼此在互联网中的位置,主机在互联网中的位置是通过ip地址标记的,当找到ip地址后,再通过端口号标识运行在主机中的进程从而实现通信. IP地址: IP地 ...

  8. linux动态库加载路径修改

    1.在 /etc/ld.so.conf 文件中添加搜索路径,重启或者 ldconfig 生效: 2.在 /etc/ld.so.conf.d 目录下添加 *.conf 文件,其中可以添加搜索路径,重启获 ...

  9. java Beanutils.copyProperties( )用法

    这是一篇开发自辩甩锅稿~~~~ 昨天测试小姐姐将我的一个bug单重开了,emmmm....内心OS:就调整下对象某个属性类型这么简单的操作,我怎么可能会出错呢,一定不是我的锅!!but再怎么抗拒,bu ...

  10. 【04】JSONP 教程

    JSONP 教程 Jsonp(JSON with Padding) 是 json 的一种"使用模式",可以让网页从别的域名(网站)那获取资料,即跨域读取数据. 为什么我们从不同的域 ...