数独的 Dancing links 解法（含源代码）

数独的 Dancing links 解法（含源代码）

 数独问题可以转换为729行324列的exact cover 问题。每一行代表每个方格的可选值，每一列代表每个格的限制，建立双向十字链表，即可用dancing links算法优化求解。

  1 Source Code
  2
  3 Problem:  3074   User: theorix 
  4 Memory: 308K  Time: 47MS 
  5 Language: C ++   Result: Accepted 
  6
  7 Source Code 
  8 #include < iostream >
  9 using   namespace  std;
 10 #define  RR 729
 11 #define  CC 324
 12 #define  INF 1000000000
 13 int  mem[RR + 9 ];
 14 int  ans[RR + 9 ];
 15 char  ch[RR + 9 ];
 16 int  cnt[CC + 9 ];
 17 struct  node
 18 {
 19    int r,c;
 20    node *up;
 21    node *down;
 22    node *left;
 23    node *right;
 24} head,all[RR * CC + 99 ],row[RR],col[CC]; int  all_t;
 25 inline  void  link( int  r, int  c)
 26 {
 27    cnt[c]++;
 28    node *t=&all[all_t++];
 29    t->r=r;
 30    t->c=c;
 31    t->left=&row[r];
 32    t->right=row[r].right;
 33    t->left->right=t;
 34    t->right->left=t;
 35    t->up=&col[c];
 36    t->down=col[c].down;
 37    t->up->down=t;
 38    t->down->up=t;
 39}
 40 inline  void  remove( int  c)
 41 {
 42    node *t,*tt;
 43    col[c].right->left=col[c].left;
 44    col[c].left->right=col[c].right;
 45    for(t=col[c].down;t!=&col[c];t=t->down)
 46    {
 47        for(tt=t->right;tt!=t;tt=tt->right)
 48        {
 49            cnt[tt->c]--;
 50            tt->up->down=tt->down;
 51            tt->down->up=tt->up;
 52        }
 53        t->left->right=t->right;
 54        t->right->left=t->left;
 55    }
 56}
 57 inline  void  resume( int  c)
 58 {
 59    node *t,*tt;
 60    for(t=col[c].down;t!=&col[c];t=t->down)
 61    {        
 62        t->right->left=t;
 63        t->left->right=t;
 64        for(tt=t->left;tt!=t;tt=tt->left)
 65        {
 66            cnt[tt->c]++;
 67            tt->down->up=tt;
 68            tt->up->down=tt;
 69        }
 70    }    
 71    col[c].left->right=&col[c];
 72    col[c].right->left=&col[c];
 73}
 74 bool  solve( int  k)
 75 {
 76    if(head.right==&head)
 77        return true;
 78    node*t,*tt;
 79    int min=INF,tc;
 80    for(t=head.right;t!=&head;t=t->right)
 81    {
 82        if(cnt[t->c]<min)
 83        {
 84            min=cnt[t->c];
 85            tc=t->c;
 86            if(min<=1)break;
 87        }
 88    }
 89    remove(tc);
 90    for(t=col[tc].down;t!=&col[tc];t=t->down)
 91    {
 92        mem[k]=t->r;
 93        t->left->right=t;
 94        for(tt=t->right;tt!=t;tt=tt->right)
 95        {
 96            remove(tt->c);
 97        }
 98        t->left->right=t->right;
 99        if(solve(k+1))
100            return true;
101        t->right->left=t;
102        for(tt=t->left;tt!=t;tt=tt->left)
103        {
104            resume(tt->c);
105        }
106        t->right->left=t->left;
107    }
108    resume(tc);
109    return false;
110}
111 int  main()
112 {
113    double ss=0;
114    while(gets(ch))
115    {
116        int i,j,k;
117        if(ch[0]=='e')break;
118        all_t=1;
119        memset(cnt,0,sizeof(cnt));
120        head.left=&head;
121        head.right=&head;
122        head.up=&head;
123        head.down=&head;
124        head.r=RR;
125        head.c=CC;
126        for(i=0;i<CC;i++)
127        {
128            col[i].c=i;
129            col[i].r=RR;
130            col[i].up=&col[i];
131            col[i].down=&col[i];
132            col[i].left=&head;
133            col[i].right=head.right;
134            col[i].left->right=&col[i];
135            col[i].right->left=&col[i];
136        }
137        for(i=0;i<RR;i++)
138        {
139            row[i].r=i;
140            row[i].c=CC;
141            row[i].left=&row[i];
142            row[i].right=&row[i];
143            row[i].up=&head;
144            row[i].down=head.down;
145            row[i].up->down=&row[i];
146            row[i].down->up=&row[i];
147        }
148        for(i=0;i<RR;i++)
149        {
150            int r=i/9/9%9;
151            int c=i/9%9;
152            int val=i%9+1;
153            if(ch[r*9+c]=='.'||ch[r*9+c]==val+'0')
154            {
155                link(i,r*9+val-1);
156                link(i,81+c*9+val-1);
157                int tr=r/3;
158                int tc=c/3;
159                link(i,162+(tr*3+tc)*9+val-1);
160                link(i,243+r*9+c);
161            }
162        }
163        for(i=0;i<RR;i++)
164        {
165            row[i].left->right=row[i].right;
166            row[i].right->left=row[i].left;
167        }
168        solve(1);
169        for(i=1;i<=81;i++)
170        {
171            int t=mem[i]/9%81;
172            int val=mem[i]%9+1;
173            ans[t]=val;
174        }
175        for(i=0;i<81;i++)
176            printf("%d",ans[i]);
177        printf("\n");
178    }
179}
180
181