• POJ 3074 Sudoku (Dacing Links)


    推荐一个写数独很好的博客:http://www.cnblogs.com/grenet/p/3163550.html

    主要是把九宫格里的元素换到矩阵里面再求解dancing links

    网上找的一模版

      1 #include<cstdio>
      2 #include<cstring>
      3 #define MAXM 10
      4 #define MAXL 324
      5 #define MAXN 240000
      6 #define INF 0x7FFFFFFF
      7 char sd[MAXM][MAXM];
      8 int L[MAXN], R[MAXN], U[MAXN], D[MAXN], H[MAXN];
      9 int size, C[MAXN], S[MAXN], X[MAXN], Q[MAXN], vis[MAXL * 3];
     10 bool Read()
     11 {
     12     int i, j;
     13     for (i = 1; i < MAXM; i++)
     14     {
     15         for (j = 1; j < MAXM; j++)
     16         {
     17             scanf(" %c", &sd[i][j]);
     18             if (sd[i][j] == 'e')
     19                 return false;
     20         }
     21     }
     22     return true;
     23 }
     24 void Init()
     25 {
     26     int i;
     27     for (i = 0; i <= MAXL; i++)
     28     {
     29         L[i + 1] = i;
     30         R[i] = i + 1;
     31         U[i] = D[i] = i;
     32         S[i] = 0;
     33     }
     34     R[MAXL] = 0;
     35     size = MAXL + 1;
     36 }
     37 void Remove(int c)
     38 {
     39     int i, j;
     40     L[R[c]] = L[c];
     41     R[L[c]] = R[c];
     42     for (i = D[c]; i != c; i = D[i])
     43     {
     44         for (j = R[i]; j != i; j = R[j])
     45         {
     46             U[D[j]] = U[j];
     47             D[U[j]] = D[j];
     48             S[C[j]]--;
     49         }
     50     }
     51 }
     52 void Resume(int c)
     53 {
     54     int i, j;
     55     L[R[c]] = c;
     56     R[L[c]] = c;
     57     for (i = D[c]; i != c; i = D[i])
     58     {
     59         for (j = R[i]; j != i; j = R[j])
     60         {
     61             U[D[j]] = j;
     62             D[U[j]] = j;
     63             S[C[j]]++;
     64         }
     65     }
     66 }
     67 inline void Link(int r, int c)
     68 {
     69     D[size] = D[c];
     70     U[size] = c;
     71     U[D[c]] = size;
     72     D[c] = size;
     73     if (H[r] < 0)
     74         H[r] = L[size] = R[size] = size;
     75     else
     76     {
     77         L[size] = H[r];
     78         R[size] = R[H[r]];
     79         L[R[H[r]]] = size;
     80         R[H[r]] = size;
     81     }
     82     S[c]++;
     83     X[size] = r;
     84     C[size++] = c;
     85 }
     86 bool Dance(int now)
     87 {
     88     int i, j, c, temp;
     89     if (R[0] == 0)
     90         return true;
     91     for (temp = INF,i = R[0]; i; i = R[i])
     92     {
     93         if (S[i] < temp)
     94         {
     95             temp = S[i];
     96             c = i;
     97         }
     98     }
     99     Remove(c);
    100     for (i = D[c]; i != c; i = D[i])
    101     {
    102         vis[X[i]] = true;
    103         for (j = R[i]; j != i; j = R[j])
    104             Remove(C[j]);
    105         if (Dance(now + 1))
    106             return true;
    107         for (j = L[i]; j != i; j = L[j])
    108             Resume(C[j]);
    109         vis[X[i]] = false;
    110     }
    111     Resume(c);
    112     return false;
    113 }
    114 int main()
    115 {
    116     int i, j, k, r;
    117     while (Read())
    118     {
    119         Init();
    120         for (r = 0, i = 1; i < MAXM; i++)
    121         {
    122             for (j = 1; j < MAXM; j++)
    123             {
    124                 if (sd[i][j] == '.')
    125                 {
    126                     for (k = 1; k < MAXM; k++)
    127                     {
    128                         H[++r] = -1;
    129                         Q[r] = k;
    130                         Link(r, (i - 1) * 9 + k);
    131                         Link(r, 81 + (j - 1) * 9 + k);
    132                         Link(r, 162 + ((i - 1) / 3 * 3 + (j - 1) / 3) * 9 + k);
    133                         Link(r, 243 + (i - 1) * 9 + j);
    134                     }
    135                 }
    136                 else
    137                 {
    138                     H[++r] = -1;
    139                     k = sd[i][j] - '0';
    140                     Q[r] = k;
    141                     Link(r, (i - 1) * 9 + k);
    142                     Link(r, 81 + (j - 1) * 9 + k);
    143                     Link(r, 162 + ((i - 1) / 3 * 3 + (j - 1) / 3) * 9 + k);
    144                     Link(r, 243 + (i - 1) * 9 + j);
    145                 }
    146             }
    147         }
    148         memset(vis, false, sizeof(vis));
    149         Dance(0);
    150         for (i = 1; i <= r; i++)
    151         {
    152             if (vis[i])
    153                 printf("%d", Q[i]);
    154         }
    155         putchar('
    ');
    156     }
    157     return 0;
    158 }
    View Code
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  • 原文地址:https://www.cnblogs.com/ITUPC/p/5311088.html
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