【ATC】:Red Polyomino 搜索+ 剪枝

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传送门

题意

给你 $n \times n$ 的单元矩阵，其中各单元属性要么是白要么是黑，现在需要你计算出有多少种方案可以将白块绘制成红块，使得这个区域连通且大小为 $k$ 。

分析

这个数据范围大概就是用 $d f s$ 去爆搜了，关键怎么去剪枝
我们可以去用 $m a p$ 存一下这个图有没有被搜索过，然后进行剪枝即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
vector<string> s;
map<vector<string>, int> M;
int ans;
int n, m;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};void dfs(int num) {if (M[s]) return;M[s]++;if (!num) {ans++;return;}for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)if (s[i][j] == '.') {for (int k = 0; k < 4; k++) {int a = i + dx[k], b = j + dy[k];if (a >= 0 && a < n && b >= 0 && b < n && s[a][b] == 'X') {s[i][j] = 'X';dfs(num - 1);s[i][j] = '.';}}}
}void solve() {for (int i = 0; i < n; i++)for (int j = 0; j < n; j++) {if (s[i][j] == '.') {s[i][j] = 'X';dfs(m - 1);s[i][j] = '.';}}
}int main() {read(n), read(m);for (int i = 0; i < n; i++) {string x;cin >> x;s.pb(x);}solve();di(ans);return 0;
}