1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4 #include<algorithm>
5 #include<cmath>
6 #include<iomanip>
7 using namespace std;
8
9 const int n=729,m=324;
10 bool mx[2000][2000];//数独转化过来的01矩阵
11 int map[10][10],cnt[2000],head,cur,ans;
12 int sqr[10][10]={{0,0,0,0,0,0,0,0,0,0}, //九宫格的编号
13 {0,1,1,1,4,4,4,7,7,7},
14 {0,1,1,1,4,4,4,7,7,7},
15 {0,1,1,1,4,4,4,7,7,7},
16 {0,2,2,2,5,5,5,8,8,8},
17 {0,2,2,2,5,5,5,8,8,8},
18 {0,2,2,2,5,5,5,8,8,8},
19 {0,3,3,3,6,6,6,9,9,9},
20 {0,3,3,3,6,6,6,9,9,9},
21 {0,3,3,3,6,6,6,9,9,9}};
22
23 int w[10][10]={{0,0,0,0,0,0,0,0,0,0},
24 {0,6,6,6,6,6,6,6,6,6},
25 {0,6,7,7,7,7,7,7,7,6},
26 {0,6,7,8,8,8,8,8,7,6},
27 {0,6,7,8,9,9,9,8,7,6},
28 {0,6,7,8,9,10,9,8,7,6},
29 {0,6,7,8,9,9,9,8,7,6},
30 {0,6,7,8,8,8,8,8,7,6},
31 {0,6,7,7,7,7,7,7,7,6},
32 {0,6,6,6,6,6,6,6,6,6}};
33
34 struct point
35 {
36 int row,lc,rc,up,down,col;//元素的上下左右，行号和列标
37 }node[2000*2000];
38
39 inline int id(int x,int y)
40 {
41 return (x-1)*9+y;//格子（x，y）的编号
42 }
43
44 void init(int c)//初始化列标元素
45 {
46 for (int i=0;i<=c;i++)
47 {
48 node[i].lc=i-1;
49 node[i].rc=i+1;
50 node[i].up=node[i].down=node[i].col=i;
51 }
52 node[0].lc=c;
53 node[c].rc=0;
54 }
55
56 void build_link()//把01矩阵中的 1 用dancing link 连接
57 {
58 cur=m;
59 for (int i=1;i<=n;i++)
60 {
61 int start,pre;
62 start=pre=cur+1;
63 for (int j=1;j<=m;j++)
64 if (mx[i][j])
65 {
66 cur++;
67 cnt[j]++;
68 node[cur].row=i;
69
70 node[cur].lc=pre;
71 node[cur].rc=start;
72 node[pre].rc=cur;
73 node[start].lc=cur;
74
75 node[cur].col=j;
76 node[cur].up=node[j].up;
77 node[cur].down=j;
78 node[node[j].up].down=cur;
79 node[j].up=cur;
80 pre=cur;
81 }
82 }
83 }
84
85 inline void cover(int c)//删除冲突元素
86 {
87 for (int i=node[c].up;i!=c;i=node[i].up)
88 for (int j=node[i].rc;j!=i;j=node[j].rc)
89 {
90 node[node[j].up].down=node[j].down;
91 node[node[j].down].up=node[j].up;
92 cnt[node[j].col]--;
93 }
94 node[node[c].lc].rc=node[c].rc;
95 node[node[c].rc].lc=node[c].lc;
96 }
97
98 inline void uncover(int c)//恢复冲突元素
99 {
100 for (int i=node[c].up;i!=c;i=node[i].up)
101 for (int j=node[i].rc;j!=i;j=node[j].rc)
102 {
103 node[node[j].up].down=j;
104 node[node[j].down].up=j;
105 cnt[node[j].col]++;
106 }
107 node[node[c].lc].rc=c;
108 node[node[c].rc].lc=c;
109 }
110
111 void read_data()//把数独转化成01矩阵
112 {
113 for (int i=1;i<=9;i++)
114 for (int j=1;j<=9;j++)
115 {
116 scanf("%d",&map[i][j]);
117 int c=id(i,j),t,k;
118 if (map[i][j])//数独中本来有数，直接加入
119 {
120 k=map[i][j];
121 t=(c-1)*9+k;
122 mx[t][c]=true;
123 mx[t][81+9*(i-1)+k]=true;
124 mx[t][162+9*(j-1)+k]=true;
125 mx[t][243+(sqr[i][j]-1)*9+k]=true;
126 }
127 else
128 {
129 for (k=1;k<=9;k++) //数独中本来没数，那么加入1-9的情况
130 {
131 t=(c-1)*9+k;
132 mx[t][c]=true;
133 mx[t][81+9*(i-1)+k]=true;
134 mx[t][162+9*(j-1)+k]=true;
135 mx[t][243+(sqr[i][j]-1)*9+k]=true;
136 }
137 }
138 }
139 }
140
141 bool dfs(int step,int score)
142 {
143 if (node[head].rc==head) //已经全部覆盖
144 {
145 ans=max(score,ans);
146 return true;
147 }
148
149 int i,j,c,t=210000,x,y,num,flag=0;
150 for (i=node[head].rc;i!=head;i=node[i].rc) //启发式，每次处理元素最少的列
151 if (cnt[i]<t)
152 {
153 t=cnt[i];
154 c=i;
155 }
156 if (t==0)
157 return false;
158
159 cover(c);//覆盖当前列
160
161 for (i=node[c].down;i!=c;i=node[i].down)
162 {
163 for (j=node[i].lc;j!=i;j=node[j].lc)//删除冲突的行
164 cover(node[j].col);
165 num=(node[i].row-1)/9+1;
166 x=(num-1)/9+1;
167 y=num-9*(x-1);
168 flag|=dfs(step+1,score+w[x][y]*(node[i].row-(num-1)*9));
169 for (j=node[i].rc;j!=i;j=node[j].rc)//恢复删除冲突的行
170 uncover(node[j].col);
171 }
172
173 uncover(c);//恢复当前列
174 return flag;
175 }
176
177 void solve()
178 {
179 init(m);
180 build_link();
181 int flag=1;
182 if (!dfs(1,0))
183 printf("-1\n");
184 else printf("%d\n",ans);
185 }
186
187 int main()
188 {
189 read_data();
190 solve();
191 return 0;
192 }
