【题解】P4460 [CQOI2018] 解锁屏幕

P4460 CQOI2018 解锁屏幕 - 洛谷 (luogu.com.cn)

本来想的是 bfs，后来发现随手就炸了，还不好控制可重状态。

改成了状态压缩 dp，判断可移动时候可以 用直线斜率判断。

#include<bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 20;
const int M = 2010;
const LL P = 100000007;
int X[N], Y[N];

int bet[N][N];
LL dp[(1 << N) + 10][N];

int main () {
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	int n; cin >> n;
	for (int i = 1; i <= n; i ++) {
		cin >> X[i] >> Y[i];
	}

	for (int i = 1; i <= n; i++) {
	    for (int j = 1; j <= n; j++) {
	        if (i == j) continue;
	        bet[i][j] = 0;
	        for (int k = 1; k <= n; k++) {
	            if (k == i || k == j) continue;
	
	            // 判断 k 是否在 i -> j 的线段上
	            bool collinear =
	                (X[k] - X[i]) * (Y[j] - Y[i]) ==
	                (Y[k] - Y[i]) * (X[j] - X[i]);
	
	            bool inBox =
	                min(X[i], X[j]) <= X[k] && X[k] <= max(X[i], X[j]) &&
	                min(Y[i], Y[j]) <= Y[k] && Y[k] <= max(Y[i], Y[j]);
	
	            if (collinear && inBox) {
	                bet[i][j] |= (1 << (k - 1));
	            }
	        }
	    }
	}
	
	LL ans = 0;
	memset(dp, 0, sizeof(dp));
	for (int i = 1; i <= n; i ++) {
		dp[1 << (i - 1)][i] = 1;
	}

	for (int mask = 0; mask < (1 << n); mask++) {
	    for (int i = 1; i <= n; i ++) {
	        if (!dp[mask][i]) continue;
	        for (int j = 1; j <= n; j ++) {
	            if (mask & (1 << (j - 1))) continue;
	            if ((bet[i][j] & mask) != bet[i][j]) continue;
	
	            int nmask = mask | (1 << (j - 1));
	            dp[nmask][j] = (dp[nmask][j] + dp[mask][i]) %P;
	        }
	    }
	}
	
	ans = 0;
	for (int mask = 0; mask < (1 << n); mask++) {
	    if (__builtin_popcount(mask) >= 4) {
	        for (int i = 1; i <= n; i ++) {
	            ans = (ans + dp[mask][i]) % P;
	        }
	    }
	}
	
	cout << ans << "\n"; 
	
	return 0;
}