Sudoku（简单DFS）

Posted 2022-06-20 wsy107316

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5547

数据比较少,直接暴力DFS,检验成立情况即可

AC代码：但是不知道为什么用scanf,printf输入输出就WA了

  1 /* */
  2 # include <iostream>
  3 # include <stdio.h>
  4 # include <string.h>
  5 # include <cstdlib>
  6 # include <cmath>
  7 # include <climits>
  8 # include <deque>
  9 # include <queue>
 10 # include <stack>
 11 # include <vector>
 12 # include <map>
 13 # include <set>
 14 # include <ctime>
 15 # include <functional>
 16 using namespace std;
 17 typedef long long ll;
 18 const int inf=0x3f3f3f3f;
 19 const int maxn=5e4+50;
 20 const ll mod=1e9+7;
 21 char ma[25][25];
 22 int vis[25];
 23 int p[25][25];
 24 int tx[20];
 25 int ty[20];
 26 int flag, tot;
 27 
 28 int check()
 29 
 30     for(int i=0; i<4; i++ )///检查每一行
 31     
 32         vis[1]=vis[2]=vis[3]=vis[4]=0;
 33         for(int j=0; j<4; j++ )
 34         
 35             if( p[i][j]==0 )
 36                 continue;
 37             if( vis[p[i][j]] )///填的数重复了不成立
 38                 return 0;
 39             vis[p[i][j]] = 1;
 40         
 41     
 42 
 43     for(int j=0; j<4; j++ )///检查每一列
 44     
 45         vis[1]=vis[2]=vis[3]=vis[4]=0;
 46         for(int i=0; i<4; i++ )
 47         
 48             if( p[i][j]==0 )
 49                 continue;
 50             if( vis[p[i][j]] )
 51                 return 0;
 52                 vis[p[i][j]] = 1;
 53         
 54     
 55 
 56     vis[1] = vis[2] = vis[3] = vis[4] = 0;
 57     for(int i=0; i<2; i++ )///检查左上角的1/4方块
 58     
 59         for(int j=0; j<2; j++ )
 60         
 61             if( p[i][j]==0 )
 62                 continue;
 63             if( vis[p[i][j]] )
 64                 return 0;
 65             vis[p[i][j]] = 1;
 66         
 67     
 68 
 69     vis[1]=vis[2]=vis[3]=vis[4]=0;
 70     for(int i=0; i<2; i++ )///检查右上角的1/4块
 71     
 72         for(int j=2; j<4; j++ )
 73         
 74             if( p[i][j]==0 )
 75                 continue;
 76             if( vis[p[i][j]] )
 77                 return 0;
 78             vis[p[i][j]] = 1;
 79         
 80     
 81 
 82     vis[1] = vis[2] = vis[3] = vis[4]=0;
 83     for(int i=2; i<4; i++ )///检查左下角1/4块
 84     
 85         for(int j=0; j<2; j++ )
 86         
 87             if( p[i][j]==0 )
 88                 continue;
 89             if( vis[p[i][j]] )
 90                 return 0;
 91             vis[p[i][j]] = 1;
 92         
 93     
 94 
 95     vis[1] = vis[2] = vis[3] = vis[4] = 0;
 96     for(int i=2; i<4; i++ )///检查右下角1/4块
 97     
 98         for(int j=2; j<4; j++ )
 99         
100             if( p[i][j]==0 )
101                 continue;
102             if( vis[p[i][j]] )
103                 return 0;
104             vis[p[i][j]] = 1;
105         
106     
107     return 1;
108 
109 
110 
111 void dfs(int x)
112 
113     if( flag )
114         return ;
115     if( x==tot )
116     
117         for(int i=0; i<4; i++ )
118         
119             for(int j=0; j<4; j++ )
120             
121                 cout<<p[i][j];
122                 //printf("%d", p[i][j]);
123             
124             cout<<endl;
125             //printf("\n");
126         
127         flag=1;
128         return ;
129     
130 
131     for(int i=1; i<=4; i++ )
132     
133         p[tx[x]][ty[x]]=i;
134         if( check())
135         
136             dfs(x+1);
137         
138         p[tx[x]][ty[x]] = 0;
139     
140 
141 
142 int main()
143 
144     int t;
145     scanf("%d", &t);
146     int k=1;
147     //getchar();
148     while( t-- )
149     
150         for(int i=0; i<4; i++ )
151         
152             for(int j=0; j<4; j++ )
153             
154                 cin>>ma[i][j];
155                 //scanf("%c", &ma[i][j]);
156             
157             //getchar();
158         
159 
160         for(int i=0; i<4; i++ )
161         
162             for(int j=0; j<4; j++ )
163             
164                 if( ma[i][j]==‘*‘ )
165                 
166                     p[i][j] = 0;
167                 
168                 else
169                     p[i][j] = ma[i][j] - ‘0‘;
170             
171         
172 
173         flag=0;
174         tot = 0;
175         for(int i=0; i<4; i++ )
176         
177             for(int j=0; j<4; j++ )
178             
179                 if( !p[i][j] )
180                 
181                     /*依次记录空节点的坐标*/
182                     tx[tot] = i;
183                     ty[tot] = j;
184                     tot++;
185                 
186             
187         
188         printf("Case #%d:\n", k++);
189         dfs(0);
190     
191     return 0;
192 

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