【题解】LuoGu1040：加分二叉树

原题传送门
区间dp
根据前序遍历的特点，对于一段区间$[l,r]$，如果选择$rt(l<=rt<=r)$作为根，那么这段区间的分数是$分{数}_{[l,rt-1]}\ast 分{数}_{[rt+1,r]}+va{l}_{rt}$
所以可以令$d{p}_{i,j}$表示$[l,r]$的答案

• $d{p}_{i,i}=va{l}_i$
• $d{p}_{i,j}=max(d{p}_{i,k-1}\ast d{p}_{k+1,r}+va{l}_k)$

同时为了输出前序遍历，在更新$dp$数组的时候顺便更新$r{t}_{i,j}$表示这一段区间的根

Code：

#include 
#define maxn 110
using namespace std;
int dp[maxn][maxn], rt[maxn][maxn], n;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void print(int l, int r){
	printf("%d ", rt[l][r]);
	if (l < rt[l][r]) print(l, rt[l][r] - 1);
	if (rt[l][r] < r) print(rt[l][r] + 1, r);
}

int main(){
	n = read();
	for (int i = 1; i <= n; ++i) dp[i][i] = read(), rt[i][i] = i;
	for (int l = 2; l <= n; ++l)
		for (int i = 1, j = i + l - 1; j <= n; ++i, ++j){
			if (dp[i + 1][j] + dp[i][i] > dp[i][j]) dp[i][j] = dp[i + 1][j] + dp[i][i], rt[i][j] = i;
			if (dp[i][j - 1] + dp[j][j] > dp[i][j]) dp[i][j] = dp[i][j - 1] + dp[j][j], rt[i][j] = j;
			for (int k = i + 1; k < j; ++k)
				if (dp[i][k - 1] * dp[k + 1][j] + dp[k][k] > dp[i][j])
					dp[i][j] = dp[i][k - 1] * dp[k + 1][j] + dp[k][k], rt[i][j] = k;
		}
	printf("%d\n", dp[1][n]);
	print(1, n);
	return 0;
}