dlx算法解数独,有空会写详解,暂时就这样了

.2738..1..1...6735.......293.5692.8...........6.1745.364.......9518...7..8..6534.
......52..8.4......3...9...5.1...6..2..7........3.....6...1..........7.4.......3.
end

527389416819426735436751829375692184194538267268174593643217958951843672782965341
416837529982465371735129468571298643293746185864351297647913852359682714128574936

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/*=============================================================================
#     FileName: 3074.cpp
#         Desc:
#       Author: zhuting
#        Email: cnjs.zhuting@gmail.com
#     HomePage: my.oschina.net/locusxt
#      Version: 0.0.1
#    CreatTime: 2014-01-16 21:44:18
#   LastChange: 2014-01-16 21:44:18
#      History:
=============================================================================*/
/** 数独需要满足的四个条件:* 1.每个格子都有数字* 2.每一行要有1-9* 3.每一列要有1-9* 4.每个3*3的块要有1-9** 1-81列 表示某个格子有没有填数字,填了则为1* 82-162列 表示某一行有没有1-9* 163-243列 表示某一列有没有1-9* 244-324列 表示某个块有没有1-9** 共9*9*9行* 表示某一个格子填了某个数后的影响*/#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#define maxl 100
#define maxc 325 //最大列数,包含头结点
#define maxr 730 //最大行数,包含待匹配行
#define maxn 240000 //最大总结点数
#define head 0 //head的编号为0
#define GET_CEL(i, j, k) i*9+j
#define GET_ROW(i, j, k) 80+i*9+k
#define GET_COL(i, j, k) 161+j*9+k
#define GET_BLO(i, j, k) 242+(3*(i/3)+j/3)*9+k
#define GET_I(l) (l-1)/81
#define GET_J(l) (l-1)%81/9
#define GET_K(l) (l-1)%81%9+1const int m = 729;//行数
const int n = 324;//列数
bool mymap[maxr][maxc];int num[maxc] = {0};//数组实现的dancing links
int up[maxn] = {0}, down[maxn] = {0}, left[maxn] = {0}, right[maxn] = {0};//顾名思义是某个结点的上下左右结点的标号
int column[maxn] = {0}, row[maxn] = {0}, ans[maxn] = {0};//某个结点对应的行号,ans是取的行号的数组char str[maxl];/** 每填一个数,假设填在数独i行j列填的是k* 那么会产生以下影响:* 9*j+i个cell被填上了数字*///生成初始的01串
void gen_zero_one()
{int cur_line = 0;for (int i = 0; i < 9; ++i){for (int j = 0; j < 9; ++j){for (int k = 1; k <= 9; ++k){mymap[cur_line][GET_CEL(i, j, k)] = 1;mymap[cur_line][GET_ROW(i, j, k)] = 1;mymap[cur_line][GET_COL(i, j, k)] = 1;mymap[cur_line][GET_BLO(i, j, k)] = 1;++cur_line;}}}return;
}//生产初始的节点表
void gen_node()
{int id = 1;int tmp = 0;int lineup[maxc] = {0};for (int j = 0; j < n; ++j)//待匹配行{lineup[j] = id;column[id] = j + 1;if (j != n - 1){right[id] = id + 1;}left[id] = id - 1;++id;}right[head] = 1;left[head] = id - 1;right[id - 1] = head;for (int i = 0; i < m; ++i){int first_id = 0;int left_id = 0;for (int j = 0; j < n; ++j){//scanf("%d", &tmp);tmp = mymap[i][j];if (tmp){row[id] = i + 1;column[id] = j + 1;if (first_id == 0){first_id = id;left_id = id;}else{left[id] = left_id;right[left_id] = id;left_id = id;}int upid = lineup[j];up[id] = upid;down[upid] = id;lineup[j] = id;++id;}}if (first_id != 0){left[first_id] = left_id;right[left_id] = first_id;}}for (int j = 0; j < n; ++j){down[lineup[j]] = j + 1;up[j + 1] = lineup[j];}//test(id);return;
}void init()//初始化
{memset(up, 0, sizeof(up));memset(down, 0, sizeof(down));memset(left, 0, sizeof(left));memset(right, 0, sizeof(right));memset(column, 0, sizeof(column));memset(row, 0, sizeof(row));memset(ans, 0, sizeof(ans));memset(mymap, 0, sizeof(mymap));for (int i = 1; i <= n; ++i)num[i] = 9;return;
}//这里的remove其实并没有真正删除掉结点,可以用resume恢复
void remove(int c)//去掉c号结点,以及在c所在列上有结点的行的结点
{left[right[c]] = left[c];right[left[c]] = right[c];for (int i = down[c]; i != c; i = down[i]){for (int j = right[i]; j != i; j = right[j]){up[down[j]] = up[j];down[up[j]] = down[j];--num[column[j]];}}return;
}void resume(int c)//remove的逆操作
{left[right[c]] = c;right[left[c]] = c;for (int i = up[c]; i != c; i = up[i]){for (int j = right[i]; j != i; j = right[j]){up[down[j]] = j;down[up[j]] = j;++num[column[j]];}}return;
}void recover(int t)
{for (int i = 0; i < t; ++i){int tmp = ans[i];int cur = GET_I(tmp) * 9 + GET_J(tmp);str[cur] = GET_K(tmp) + '0';/*printf("%d:\n", tmp);printf("%d %d %d\n", GET_I(tmp), GET_J(tmp), GET_K(tmp));*/}return;
}//我们的终极目标实际上是把十字链表中所有的结点都删掉,只留一个头结点
bool dance(int k)//这里的k实际上是没多大意义的
{//printf("dance: %d\n", k);int c = right[head];//c是头结点的右边一个结点if (c == head)//只剩下十字链表的头结点,则完成了目标{recover(k);return true;}int min_n = num[c];int min_cur = c;while(true){c = right[c];if (c == head)break;if(num[c] < min_n){min_n = num[c];min_cur = c;}}c = min_cur;remove(c);for (int i = down[c]; i != c; i = down[i])//这里其实是枚举某一行对待匹配行的第c位进行匹配{ans[k] = row[i];for (int j = right[i]; j != i; j = right[j])remove(column[j]);//remove待匹配行的column[j]结点if (dance(k + 1))return true;for (int j = left[i]; j != i; j = left[j])resume(column[j]);}resume(c);return false;
}void parse(char* a)
{int tmp = 0;for(int i = 0; i < 9; ++i){for (int j = 0; j < 9; ++j){if(a[i * 9 + j] != '.'){tmp = a[i * 9 + j] - '0';remove(GET_CEL(i, j, tmp) + 1);remove(GET_ROW(i, j, tmp) + 1);remove(GET_COL(i, j, tmp) + 1);remove(GET_BLO(i, j, tmp) + 1);}}}return;
}int main()
{while (true){scanf("%s", str);if (str[0] == 'e'){break;}init();gen_zero_one();gen_node();parse(str);int is_possible = dance(0);if (is_possible){printf("%s\n", str);}elseprintf("error\n");}return 0;
}