poj3074 Sudoku--舞蹈链数独

原题链接：http://poj.org/problem?id=3074

题意：给定一个9*9的数独，求解。

#define _CRT_SECURE_NO_DEPRECATE #include<iostream>#include<vector>#include<cstring>#include<queue>#include<stack>#include<algorithm>#include<cmath>#define INF 99999999#define eps 0.0001const int MAXC = 81 * 4 + 10;const int MAXR = 81 * 9 + 10;const int MAXN = MAXC*MAXR + 10;using namespace std;int cnt;char s[200];int U[MAXN], D[MAXN], L[MAXN], R[MAXN];int C[MAXN], M[MAXN];//节点所在列与行int H[MAXR];//行头指针int S[MAXC];//储存每列的元素数量int ans[100];void init(int n){for (int i = 0; i <= n; i++){U[i] = D[i] = i;L[i] = i - 1;R[i] = i + 1;S[i] = 0;}R[n] = 0;L[0] = n;S[0] = INF + 1;////////////////cnt = n + 1;memset(H, 0, sizeof(H));}void link(int r, int c){M[cnt] = r;//记录行列C[cnt] = c;U[cnt] = U[c];//上下连接D[cnt] = c;D[U[c]] = cnt;U[c] = cnt;if (H[r] == 0)//左右连接H[r] = L[cnt] = R[cnt] = cnt;else{L[cnt] = L[H[r]];R[L[H[r]]] = cnt;L[H[r]] = cnt;R[cnt] = H[r];}S[c]++;cnt++;}void remove(int c){L[R[c]] = L[c];R[L[c]] = R[c];for (int i = D[c]; i != c; i = D[i]){for (int j = R[i]; j != i; j = R[j]){U[D[j]] = U[j];D[U[j]] = D[j];S[C[j]]--;}}}void resume(int c){L[R[c]] = c;R[L[c]] = c;for (int i = U[c]; i != c; i = U[i]){for (int j = L[i]; j != i; j = L[j]){U[D[j]] = j;D[U[j]] = j;S[C[j]]++;}}}bool dance(int k){if (R[0] == 0){sort(ans, ans + 81);int p = 0;for (int i = 0; i < 9; i++){for (int j = 0; j < 9; j++){int num = ans[p++];num = num - (i * 9 + j) * 9;printf("%d", num);}}printf("\n");return true;}int c;int t = INF;for (int i = R[0]; i != 0; i = R[i]){if (S[i] < t){t = S[i];c = i;}}remove(c);for (int i = D[c]; i != c; i = D[i]){ans[k] = M[i];for (int j = R[i]; j != i; j = R[j])remove(C[j]);if (dance(k + 1))return true;for (int j = L[i]; j != i; j = L[j])resume(C[j]);}resume(c);return false;}void build(){int p = 0;init(9 * 9 * 4);for (int i = 0; i < 9; i++){for (int j = 1; j <= 9; j++, p++){int base = (i * 9 + j - 1) * 9;if (s[p] == '.'){for (int k = 1; k <= 9; k++){int r = base + k;link(r, i * 9 + k);//第i行有数字klink(r, 81 + (j - 1) * 9 + k);//第j列有数字klink(r, 81 * 2 + ((j - 1) / 3 * 3 + i / 3) * 9 + k);//第k块有数字klink(r, 81 * 3 + i * 9 + j);//第i行j列有一个数字（限制一个格子只填一个数）}}else{int k = s[p] - '0';int r = base + k;link(r, i * 9 + k);//第i行有数字klink(r, 81 + (j - 1) * 9 + k);//第j列有数字klink(r, 81 * 2 + ((j - 1) / 3 * 3 + i / 3) * 9 + k);//第k块有数字klink(r, 81 * 3 + i * 9 + j);//第i行j列有一个数字（限制一个格子只填一个数）}}}}int main(){while (~scanf("%s", s)){if (strcmp(s, "end") == 0)break;build();dance(0);}return 0;}