POJ 3074 3076 Sudoku（Dancing Links）
题意：
求解数独。
思路：
1. 如何用 dancing links 求解数独？其中的精确覆盖策略十分精妙，感叹之余只能膜拜 Knuth 了；
2. 如何把数独问题分解成矩阵，假设是 9*9 数独问题，则分下面 4 种限制条件：
a) 每一行元素不能重复;
b) 每一列元素不能重复;
c) 每个 3*3 矩阵元素不能重复;
d) 每个格子只能存放一个数字;
3. 达到上述 4 个条件之后就可以构造 01 方阵了，列表示对于元素的限制，行代表所有格子的选择。所以有 81 * 4 列（对应上面 4 个条件），有 81 * 9 行。
5. 求解的要求是：选定其中的行，当且仅当把所有的列选择完毕之后，停止递归，当然要求每一列只有一个 1 存在。此时 Q 里面存放的就是一个可行解。
6. 其中 S 数组里面存放的是每一列 1 的个数。按照 DLX 算法的要求，每次递归总选择 1 最少的列开始，这样能够有效的减少分支。因为最终只是寻找一个可行解就返回。
7. Dancing Links 在下面 CSDN 链接的几篇博客比较清楚。简单描述下：
a) 选定 1 最少的列，对其进行删除，同时删除此列上存在 1 的行。因为某个列只能选择一个 1 的缘故，所以选择了此列上某个 1 便不能选择其它 1;
b) 选定删除列的某一行，删除此行上面存在 1 的列，即某一行达到了列所在的几个限制条件，理应对列进行删除。删除列时的操作同 a) 的操作;
c) 如果所有的列都被删除了，则我们的任务完成，Q 中保存的即是我们曾经选中的行，即可行解。否则重复 a) 的操作或者进行回溯操作;
http://www.cppblog.com/notonlysuccess/archive/2009/07/10/89701.aspx
http://blog.csdn.net/mu399/article/details/7627763
// 3074
#include <iostream> #include <algorithm> using namespace std; const int INFS = 0x3fffffff; const int MAXC = 81 * 4; const int MAXR = 81 * 9; const int MAXN = MAXC * MAXR + 10; int U[MAXN], D[MAXN], L[MAXN], R[MAXN]; int grid[MAXR+10][MAXC+10], col[MAXN], row[MAXN], S[MAXC+10], Q[MAXR+10], QC; 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[col[j]] -= 1; } } } void resume(int c) { for (int i = U[c]; i != c; i = U[i]) { for (int j = L[i]; j != i; j = L[j]) { U[D[j]] = D[U[j]] = j; S[col[j]] += 1; } } R[L[c]] = L[R[c]] = c; } bool dfs() { if (R[0] == 0) return true; int mins = INFS, c = 0; for (int i = R[0]; i != 0; i = R[i]) { if (S[i] < mins) mins = S[i], c = i; } remove(c); for (int i = D[c]; i != c; i = D[i]) { for (int j = R[i]; j != i; j = R[j]) remove(col[j]); Q[QC++] = row[i]; if (dfs()) return true; QC -= 1; for (int j = L[i]; j != i; j = L[j]) resume(col[j]); } resume(c); return false; } void solve(int ans[]) { QC = 0; dfs(); for (int i = 0; i < QC; i++) ans[(Q[i]-1)/9] = (Q[i] - 1) % 9 + 1; } void buildmatrix(char data[]) { for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { if (data[i*9+j] == '.') { for (int k = 1; k <= 9; k++) { int r = (i * 9 + j) * 9 + k; int c1 = 81 * 0 + i * 9 + k; int c2 = 81 * 1 + j * 9 + k; int c3 = 81 * 2 + (i / 3 * 3 + j / 3) * 9 + k; int c4 = 81 * 3 + i * 9 + j + 1; grid[r][c1] = grid[r][c2] = grid[r][c3] = grid[r][c4] = 1; } } else { int k = data[i*9+j] - '0'; int r = (i * 9 + j) * 9 + k; int c1 = 81 * 0 + i * 9 + k; int c2 = 81 * 1 + j * 9 + k; int c3 = 81 * 2 + (i / 3 * 3 + j / 3) * 9 + k; int c4 = 81 * 3 + i * 9 + j + 1; grid[r][c1] = grid[r][c2] = grid[r][c3] = grid[r][c4] = 1; } } } } void builddlx() { memset(S, 0, sizeof(S)); for (int i = 1; i <= MAXC; i++) R[i-1] = L[i+1] = U[i] = D[i] = i; R[MAXC] = L[1] = 0; int c = MAXC; for (int i = 1; i <= MAXR; i++) { int cflag = -1; for (int j = 1; j <= MAXC; j++) { if (grid[i][j] == 1) { c += 1; S[j] += 1; row[c] = i; col[c] = j; U[c] = U[j]; D[c] = j; D[U[j]] = c; U[j] = c; if (cflag == -1) { R[c] = L[c] = c; cflag = c; } else { L[c] = L[cflag]; R[c] = cflag; R[L[c]] = c; L[cflag] = c; } } } } } bool initdata() { char data[100]; scanf("%s", data); if (data[0] == 'e') return false; memset(grid, 0, sizeof(grid)); buildmatrix(data); builddlx(); return true; } int main() { while (initdata()) { int ans[100]; solve(ans); for (int i = 0; i < 81; i++) printf("%d", ans[i]); printf("\n"); } return 0; }
// 3076
#include <iostream>
#include <algorithm>
using namespace std;
const int INFS = 0x3fffffff;
const int MAXC = 16 * 16 * 4;
const int MAXR = 16 * 16 * 16;
const int MAXN = MAXC * MAXR + 10;
int U[MAXN], D[MAXN], L[MAXN], R[MAXN];
int grid[MAXR+10][MAXC+10], col[MAXN], row[MAXN], S[MAXC+10], Q[MAXR+10], QC;
char data[20][20];
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[col[j]] -= 1;
}
}
}
void resume(int c) {
for (int i = U[c]; i != c; i = U[i]) {
for (int j = L[i]; j != i; j = L[j]) {
U[D[j]] = D[U[j]] = j;
S[col[j]] += 1;
}
}
R[L[c]] = L[R[c]] = c;
}
bool dfs() {
if (R[0] == 0)
return true;
int mins = INFS, c = 0;
for (int i = R[0]; i != 0; i = R[i]) {
if (S[i] < mins)
mins = S[i], c = i;
}
remove(c);
for (int i = D[c]; i != c; i = D[i]) {
for (int j = R[i]; j != i; j = R[j])
remove(col[j]);
Q[QC++] = row[i];
if (dfs())
return true;
QC -= 1;
for (int j = L[i]; j != i; j = L[j])
resume(col[j]);
}
resume(c);
return false;
}
void solve(char ans[]) {
QC = 0;
dfs();
for (int i = 0; i < QC; i++)
ans[(Q[i]-1)/16] = (Q[i] - 1) % 16 + 'A';
}
void buildmatrix() {
for (int i = 0; i < 16; i++) {
for (int j = 0; j < 16; j++) {
if (data[i][j] == '-') {
for (int k = 1; k <= 16; k++) {
int r = (i * 16 + j) * 16 + k;
int c1 = 256 * 0 + i * 16 + k;
int c2 = 256 * 1 + j * 16 + k;
int c3 = 256 * 2 + (i / 4 * 4 + j / 4) * 16 + k;
int c4 = 256 * 3 + i * 16 + j + 1;
grid[r][c1] = grid[r][c2] = grid[r][c3] = grid[r][c4] = 1;
}
} else {
int k = data[i][j] - 'A' + 1;
int r = (i * 16 + j) * 16 + k;
int c1 = 256 * 0 + i * 16 + k;
int c2 = 256 * 1 + j * 16 + k;
int c3 = 256 * 2 + (i / 4 * 4 + j / 4) * 16 + k;
int c4 = 256 * 3 + i * 16 + j + 1;
grid[r][c1] = grid[r][c2] = grid[r][c3] = grid[r][c4] = 1;
}
}
}
}
void builddlx() {
memset(S, 0, sizeof(S));
for (int i = 1; i <= MAXC; i++)
R[i-1] = L[i+1] = U[i] = D[i] = i;
R[MAXC] = L[1] = 0;
int c = MAXC;
for (int i = 1; i <= MAXR; i++) {
int cflag = -1;
for (int j = 1; j <= MAXC; j++) {
if (grid[i][j] == 1) {
c += 1;
S[j] += 1;
row[c] = i;
col[c] = j;
U[c] = U[j];
D[c] = j;
D[U[j]] = c;
U[j] = c;
if (cflag == -1) {
R[c] = L[c] = c;
cflag = c;
} else {
L[c] = L[cflag];
R[c] = cflag;
R[L[c]] = c;
L[cflag] = c;
}
}
}
}
}
bool initdata() {
while (scanf("%s", data[0]) != EOF)
{
for (int i = 1; i < 16; i++)
scanf("%s", data[i]);
memset(grid, 0, sizeof(grid));
buildmatrix();
builddlx();
return true ;
}
return false;
}
int main() {
while (initdata()) {
char ans[300];
solve(ans);
for (int i = 0; i < 16; i++) {
for (int j = 0; j < 16; j++) {
printf("%c", ans[i*16+j]);
}
printf("\n");
}
printf("\n");
}
return 0;
}
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