dancing link模板

  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 }