poj3074-舞蹈链

题目链接：http://poj.org/problem?id=3074

Description
In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids.

. 2 7 3 8 . . 1 .
. 1 . . . 6 7 3 5
. . . . . . . 2 9
3 . 5 6 9 2 . 8 .
. . . . . . . . .
. 6 . 1 7 4 5 . 3
6 4 . . . . . . .
9 5 1 8 . . . 7 .
. 8 . . 6 5 3 4 .
Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.

Input
The input test file will contain multiple cases. Each test case consists of a single line containing 81 characters, which represent the 81 squares of the Sudoku grid, given one row at a time. Each character is either a digit (from 1 to 9) or a period (used to indicate an unfilled square). You may assume that each puzzle in the input will have exactly one solution. The end-of-file is denoted by a single line containing the word “end”.

Output
For each test case, print a line representing the completed Sudoku puzzle.

Sample Input
.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

Sample Output
527389416819426735436751829375692184194538267268174593643217958951843672782965341
416837529982465371735129468571298643293746185864351297647913852359682714128574936

题意
求解数独

思路
直接bfs好像要t(没试过..)，舞蹈链精确覆盖的经典题目，一个数独每个位置的要求，可以得到以下四个约束
1.每个位置有且只有一个数字
2.每个位置的数字在一行只能出现一次
3.每个位置的数字在一列只能出现一次
4.每个位置的数字在一个宫格内只能出现一次
然后针对每个位置可以建立舞蹈链了
前81列，为1条件的约束
82-162列，为2条件的约束
163-243列，为3条件的约束
244-324列，为4条件的约束
则舞蹈链行数为确定的点数+未确定的点数9，列数为324。
如果i行j列为数k，对应行则在(i - 1) 9 + j列，(i - 1) 9 + k + 81列， (j - 1) 9 + k + 162列，(((i - 1) / 3) 3 + ((j + 2) / 3) - 1) 9 + k + 243列建立。
如果i行j列为未知数,则k为1-9，对于每个k都应该建立行列关系。

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#include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long ll; //最大行数 const int MN = 1010; //最大列数 const int MM = 500; //最大点数 const int MNN = MN * MM + 10; struct DLX { //一共n行m列，s个节点 int n, m, size; //交叉十字链表组成部分 //第i个节点的上U下D左L右R，所在位置row行col列 int U[MNN], D[MNN], L[MNN], R[MNN], row[MNN], col[MNN]; //H数组记录行选择指针，S数组记录覆盖个数 int H[MNN], S[MM]; //res记录行个数，ans数组记录可行解 int ansd, ans[MN]; //初始化空表 void init(int x, int y) { n = x, m = y; //其中0节点作为head节点,其他作为列首节点 for (int i = 0; i <= m; ++i) { U[i] = D[i] = i; L[i] = i - 1; R[i] = i + 1; } R[m] = 0; L[0] = m; size = m; memset(S, 0, sizeof(S)); memset(H, -1, sizeof(H)); } void Link(int r, int c) { //节点数加一，设置s节点所处位置，以及S列覆盖个数加一 size++; row[size] = r; col[size] = c; S[c]++; //将s节点插入对应列中 D[size] = D[c]; U[D[c]] = size; U[size] = c; D[c] = size; if (H[r] < 0) {//如果该行没有元素，H[r]标记该行起始节点 H[r] = L[size] = R[size] = size; } else { //将该节点插入该行第一个节点后面 R[size] = R[H[r]]; L[R[H[r]]] = size; L[size] = H[r]; R[H[r]] = size; } } //精确覆盖 void remove(int c) { //删除c列 L[R[c]] = L[c]; R[L[c]] = R[c]; //删除该列上的元素对应的行 for (int i = D[c]; i != c; i = D[i]) {//枚举该列元素 for (int j = R[i]; j != i; j = R[j]) {//枚举列的某个元素所在行遍历 U[D[j]] = U[j]; D[U[j]] = D[j]; //将该列上的S数组减一 --S[col[j]]; } } } void resume(int c) { //恢复c列 for (int i = U[c]; i != c; i = U[i]) {//枚举该列元素 for (int j = L[i]; j != i; j = L[j]) { U[D[j]] = j; D[U[j]] = j; ++S[col[j]]; } } L[R[c]] = c; R[L[c]] = c; } bool dancing(int deep) { //if (ansd < deep) return false; //当矩阵为空时,说明找到一个可行解，算法终止 if (R[0] == 0) { //ansd = min(ansd, deep); ansd = deep; return true; } //找到节点数最少的列，枚举这列上的所有行 int c = R[0]; for (int i = R[0]; i != 0; i = R[i]) { if (S[i] < S[c]) { c = i; } } //删除节点数最少的列 remove(c); for (int i = D[c]; i != c; i = D[i]) { //将行r放入当前解 ans[deep] = row[i]; //行上节点对应的列上进行删除 for (int j = R[i]; j != i; j = R[j]) remove(col[j]); //进入下一层 if (dancing(deep + 1))return true; //对行上的节点对应的列进行恢复 for (int j = L[i]; j != i; j = L[j]) resume(col[j]); } //恢复节点数最少列 resume(c); return false; } //重复覆盖 //将列与矩阵完全分开 void Remove(int c) { for (int i = D[c]; i != c; i = D[i]) { L[R[i]] = L[i]; R[L[i]] = R[i]; } } void Resume(int c) { for (int i = D[c]; i != c; i = D[i]) { L[R[i]] = R[L[i]] = i; } } int vis[MNN]; //估价函数，模拟删除列，H()，函数返回的是至少还需要多少行才能完成重复覆盖 int A() { int dis = 0; for (int i = R[0]; i != 0; i = R[i]) vis[i] = 0; for (int i = R[0]; i != 0; i = R[i]) { if (!vis[i]) { dis++; vis[i] = 1; for (int j = D[i]; j != i; j = D[j]) { for (int k = R[j]; k != j; k = R[k]) { vis[col[k]] = 1; } } } } return dis; } void dance(int deep) { if (!R[0]) { //cout << res << endl; ansd = deep; return; } if (deep + A() >= ansd) return; int c = R[0]; for (int i = R[0]; i != 0; i = R[i]) { if (S[i] < S[c]) { c = i; } } for (int i = D[c]; i != c; i = D[i]) { //每次将第i列其他节点删除，只保留第i节点，为了找该行的节点 Remove(i); //将列上的节点完全与矩阵脱离，只删列首节点是不行的 for (int j = R[i]; j != i; j = R[j]) { Remove(j); } dance(deep + 1); for (int j = L[i]; j != i; j = L[j]) { Resume(j); } Resume(i); } } }; DLX dzl; int mp[15][15]; int x[1010]; int y[1010]; int kk[1010]; int main() { char s[100]; while (cin >> s) { if (strcmp(s, "end") == 0) break; for (int i = 0; i < 81; i++) { int xx = i / 9; int yy = i % 9; if (s[i] == '.') mp[xx + 1][yy + 1] = 0; else mp[xx + 1][yy + 1] = s[i] - '0'; } dzl.init(800, 324); int len = 1; for (int i = 1; i <= 9; i++) { for (int j = 1; j <= 9; j++) { if (mp[i][j] == 0) { for (int k = 1; k <= 9; k++) { dzl.Link(len, (i - 1) * 9 + j); dzl.Link(len, (i - 1) * 9 + k + 81); dzl.Link(len, (j - 1) * 9 + k + 162); dzl.Link(len, (((i - 1) / 3) * 3 + ((j + 2) / 3) - 1) * 9 + k + 243); x[len] = i, y[len] = j, kk[len] = k; len++; } } else { int k = mp[i][j]; dzl.Link(len, (i - 1) * 9 + j); dzl.Link(len, (i - 1) * 9 + k + 81); dzl.Link(len, (j - 1) * 9 + k + 162); dzl.Link(len, (((i - 1) / 3) * 3 + ((j + 2) / 3) - 1) * 9 + k + 243); x[len] = i, y[len] = j, kk[len] = k; len++; } } } dzl.dancing(0); //cout << dzl.ansd << endl; for (int i = 0; i < dzl.ansd; i++) { int r = dzl.ans[i]; mp[x[r]][y[r]] = kk[r]; } for (int i = 1; i <= 9; i++) { for (int j = 1; j <= 9; j++) { cout << mp[i][j]; } } cout << endl; } //system("pause"); }

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