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 }