n皇后学习
回溯
#include<cstdio>
const int N = 13;
int n, cnt, ans[N];
void dfs(int row) {
if (row == n) {
if (cnt < 3) {
for (int i = 0; i < n; ++i) {
if (i > 0)printf(" ");
printf("%d", ans[i] + 1);
}
printf("\n");
}
++cnt;
return;
}
for (int i = 0; i < n; ++i) {
bool flag = true;
for (int j = 0; j < row; ++j) {
if (ans[j] == i || row - j == i - ans[j] || j - row == i - ans[j]) {
flag = false;
break;
}
}
if (flag) {
ans[row] = i;
dfs(row + 1);
}
}
}
int main() {
scanf("%d", &n);
dfs(0);
printf("%d\n", cnt);
return 0;
}
但是最后13的时候会T
优化1
简单来说就是哪列哪一对角线被占了先缓存一下,到时候直接查表
#include<cstdio>
const int N = 13;
bool col[N], sub_diag[2 * N - 1], main_diag[2 * N - 1];
int n, cnt, ans[N];
void dfs(int row) {
if (row == n) {
if (cnt < 3) {
for (int i = 0; i < n; ++i) {
if (i > 0)printf(" ");
printf("%d", ans[i] + 1);
}
printf("\n");
}
++cnt;
return;
}
for (int col_idx = 0; col_idx < n; ++col_idx) {
int sub_diag_idx = col_idx + row, main_diag_idx = col_idx - row + n - 1;
if (col[col_idx] || sub_diag[sub_diag_idx] || main_diag[main_diag_idx])continue;
col[col_idx] = sub_diag[sub_diag_idx] = main_diag[main_diag_idx] = true;
ans[row] = col_idx;
dfs(row + 1);
col[col_idx] = sub_diag[sub_diag_idx] = main_diag[main_diag_idx] = false;
}
}
int main() {
scanf("%d", &n);
dfs(0);
printf("%d\n", cnt);
return 0;
}
位运算
其实就是优化1改成用位运算
#include<cstdio>
const int N = 13;
int n, cnt, ans[N], col, sub_diag, main_diag;
void dfs(int row) {
if (row == n) {
if (cnt < 3) {
for (int i = 0; i < n; ++i) {
if (i > 0)printf(" ");
printf("%d", ans[i] + 1);
}
printf("\n");
}
++cnt;
return;
}
for (int col_idx = 0; col_idx < n; ++col_idx) {
int sub_diag_idx = col_idx + row, main_diag_idx = col_idx - row + n - 1;
if (((col >> col_idx) | (sub_diag >> sub_diag_idx) | (main_diag >> main_diag_idx)) & 1)continue;
col ^= 1 << col_idx;
sub_diag ^= 1 << sub_diag_idx;
main_diag ^= 1 << main_diag_idx;
ans[row] = col_idx;
dfs(row + 1);
col ^= 1 << col_idx;
sub_diag ^= 1 << sub_diag_idx;
main_diag ^= 1 << main_diag_idx;
}
}
int main() {
scanf("%d", &n);
dfs(0);
printf("%d\n", cnt);
return 0;
}
位运算2
每一行的副对角线:row->row+n-1
所以只要右移row位,就能找到能一列被占用了
每一行主对角线:-row+n-1->-row+n-1+n-1
所以只要右移-row+n-1位,就能找到能一列被占用了
col | (sub_diag >> row) | (main_diag >> (n - 1 - row))就是被占用的列
~(col | (sub_diag >> row) | (main_diag >> (n - 1 - row)))就是可以用的列(当然还包括一些超过n的列)
#include<cstdio>
const int N = 13;
int n, cnt, ans[N], col, sub_diag, main_diag;
void dfs(int row) {
if (row == n) {
if (cnt < 3) {
for (int i = 0; i < n; ++i) {
if (i > 0)printf(" ");
printf("%d", ans[i] + 1);
}
printf("\n");
}
++cnt;
return;
}
int available = ((1 << n) - 1) & ~(col | (sub_diag >> row) | (main_diag >> (n - 1 - row)));
while (available) {
int p = available & (-available);
available ^= p;
col ^= p;
sub_diag ^= p << row;
main_diag ^= p << (n - 1 - row);
for (int i = 0; i < n; ++i) {
if ((1 << i) == p) {
ans[row] = i;
break;
}
}
dfs(row + 1);
col ^= p;
sub_diag ^= p << row;
main_diag ^= p << (n - 1 - row);
}
}
int main() {
scanf("%d", &n);
dfs(0);
printf("%d\n", cnt);
return 0;
}
位运算3
col的第i位代表这一行第i列是否被占用
sub_diag的第i位代表这一行第i个副对角线是否被占用
main_diag的第i位代表这一行第i个主对角线是否被占用
dfs到下一层的时候,col不变
在第i行,假设第k个副对角线被占用了
那么在i+1行,第k-1个副对角线就被占用了
同理
在第i行,假设第k个主对角线被占用了
那么在i+1行,第k+1个主对角线就被占用了
#include<cstdio>
const int N = 13;
int n, cnt, ans[N];
void dfs(int row, int col, int sub_diag, int main_diag) {
if (row == n) {
if (cnt < 3) {
for (int i = 0; i < n; ++i) {
if (i > 0)printf(" ");
printf("%d", ans[i] + 1);
}
printf("\n");
}
++cnt;
return;
}
int available = ((1 << n) - 1) & ~(col | sub_diag | main_diag);
while (available) {
int p = available & (-available);
available ^= p;
for (int i = 0; i < n; ++i) {
if ((1 << i) == p) {
ans[row] = i;
break;
}
}
dfs(row + 1, col | p, (sub_diag | p) >> 1, (main_diag | p) << 1);
}
}
int main() {
scanf("%d", &n);
dfs(0, 0, 0, 0);
printf("%d\n", cnt);
return 0;
}
舞蹈链
https://blog.csdn.net/qq_39942341/article/details/126681667?spm=1001.2014.3001.5501
行:
n
∗
n
n*n
n∗n个格子
列:
[
1
,
n
]
\left[1,n\right]
[1,n]用来记录这个数字填在了哪行
[
n
+
1
,
2
n
]
\left[n+1,2n\right]
[n+1,2n]用来记录这个数字填在了哪列
[
2
n
+
1
,
4
n
−
1
]
\left[2n+1,4n-1\right]
[2n+1,4n−1]用来记录这个数字填在了哪条副对角线(从
(
0
,
0
)
(0,0)
(0,0)开始)
[
4
n
,
6
n
−
2
]
\left[4n,6n-2\right]
[4n,6n−2]用来记录这个数字填在了哪条主对角线(从
(
n
−
1
,
0
)
(n-1,0)
(n−1,0)开始)
但是注意一点,对角线是无法完全覆盖的
比如下面的图,第3条副对角线就没有覆盖
所以如果行和列覆盖完了就可以返回了
#include<cstdio>
#include<cstring>
#include<algorithm>
const int H = 13;
const int M = H * H, N = 6 * H - 2;
int n, board_cnt;
struct A {
int a[H];
}board[100000];
bool cmp(const A& a, const A& b) {
int i = 0;
while (i < n && a.a[i] == b.a[i])++i;
return a.a[i] < b.a[i];
};
class DLX {
public:
static const int MAXN = M * 4 + N + 5;
int left[MAXN], right[MAXN], up[MAXN], down[MAXN];
int row[MAXN], col[MAXN], head[M+5], col_size[N+5], cnt;
int ans, ans_row[H];
DLX() :cnt(0), ans(0) {}
void init(const int& N) {
cnt = ans = 0;
for (int i = 0; i <= N; ++i) {
left[i] = i - 1;
right[i] = i + 1;
up[i] = down[i] = i;
}
left[0] = N;
right[N] = 0;
memset(head, -1, sizeof(head));
memset(col_size, 0, sizeof(col_size));
cnt = N + 1;
}
void link(const int& r, const int& c) {
++col_size[c];
row[cnt] = r;
col[cnt] = c;
up[cnt] = c;
down[cnt] = down[c];
up[down[c]] = cnt;
down[c] = cnt;
if (head[r] == -1) {
head[r] = left[cnt] = right[cnt] = cnt;
}
else {
right[cnt] = head[r];
left[cnt] = left[head[r]];
right[left[head[r]]] = cnt;
left[head[r]] = cnt;
}
++cnt;
}
void remove(const int& c) {
right[left[c]] = right[c];
left[right[c]] = left[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];
--col_size[col[j]];
}
}
}
void resume(const int& 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;
++col_size[col[j]];
}
}
right[left[c]] = c;
left[right[c]] = c;
}
void dance(int dep) {
if (right[0]>n) {
for (int i = 0; i < n; ++i) {
int x = (ans_row[i] - 1) / n;
int y = (ans_row[i] - 1) % n + 1;
board[board_cnt].a[x] = y;
}
++board_cnt;
return;
}
int c = right[0];
for (int i = right[c]; i != 0 && i <= n; i = right[i]) {
if (col_size[i] < col_size[c]) {
c = i;
}
}
remove(c);
for (int i = down[c]; i != c; i = down[i]) {
ans_row[ans++] = row[i];
for (int j = right[i]; j != i; j = right[j]) {
remove(col[j]);
}
dance(dep + 1);
for (int j = left[i]; j != i; j = left[j]) {
resume(col[j]);
}
--ans;
}
resume(c);
return;
}
};
int main() {
scanf("%d", &n);
DLX solver;
solver.init(6 * n - 2);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
int idx = i * n + j + 1;
solver.link(idx, i + 1);
solver.link(idx, n + j + 1);
solver.link(idx, 2 * n + 1 + i + j);
solver.link(idx, 4 * n + n + j - i - 1);
}
}
solver.dance(0);
std::sort(board, board + board_cnt, cmp);
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < n; ++j) {
if (j > 0)printf(" ");
printf("%d", board[i].a[j]);
}
printf("\n");
}
printf("%d\n", board_cnt);
return 0;
}