\(\mathcal{Description}\)

  Link.

  給定一棵 \(n\) 層的完全二叉樹，你把每個結點染成黑色或白色，滿足黑色葉子個數不超過 \(m\)。對於一個葉子 \(u\)，若其 \(k\) 級父親與其同為黑色，則對答案貢獻 \(a_{uk}\)；若同為白色，則對答案貢獻 \(b_{uk}\)。求最大貢獻和。

  \(n\le10\)。

\(\mathcal{Solution}\)

  想要 DP，比如令 \(f(u,i)\) 表示 \(u\) 子樹內有 \(i\) 個葉子為黑色時的最大貢獻和。但發現這根本沒法轉移 qwq。

  那……爆搜呢？

  從上往下搜尋，直接欽定當前非葉結點是黑是白，搜到葉子時，在向上計算當前葉子是黑色/白色時的貢獻，回溯時簡單揹包。複雜度 \(\mathcal O(n4^n)\)

\(\mathcal{Code}\)

/* Clearink */

#include <cstdio>

const int MAXN = 10;
int n, m, a[1 << MAXN | 5][MAXN + 5], b[1 << MAXN | 5][MAXN + 5];
int f[1 << MAXN | 5][1 << MAXN | 5];
bool fight[1 << MAXN | 5];

inline void chkmax ( int& a, const int b ) { a < b && ( a = b, 0 ); }

inline void solve ( const int u, const int d ) {
	for ( int i = 0; i <= 1 << d; ++ i ) f[u][i] = 0;
	if ( !d ) {
		for ( int i = 1; i <= n; ++ i ) {
			if ( fight[u >> i] ) f[u][1] += a[u][i];
			else f[u][0] += b[u][i];
		}
	} else {
		for ( int k = 0; k <= 1; ++ k ) {
			fight[u] = k;
			solve ( u << 1, d - 1 ), solve ( u << 1 | 1, d - 1 );
			for ( int i = 0; i <= 1 << d >> 1; ++ i ) {
				for ( int j = 0; j <= 1 << d >> 1; ++ j ) {
					chkmax ( f[u][i + j], f[u << 1][i] + f[u << 1 | 1][j] );
				}
			}
		}
	}
}

int main () {
	scanf ( "%d %d", &n, &m ), -- n;
	for ( int i = 0; i < 1 << n; ++ i ) {
		for ( int j = 1; j <= n; ++ j ) {
			scanf ( "%d", &a[( 1 << n ) + i][j] );
		}
	}
	for ( int i = 0; i < 1 << n; ++ i ) {
		for ( int j = 1; j <= n; ++ j ) {
			scanf ( "%d", &b[( 1 << n ) + i][j] );
		}
	}
	solve ( 1, n );
	int ans = 0;
	for ( int i = 0; i <= m; ++ i ) chkmax ( ans, f[1][i] );
	printf ( "%d\n", ans );
	return 0;
}