【题解】LuoGu2015：二叉苹果树

原题传送门
这应该是树型dp的例题，我竟然才做
现在的水平直接秒杀掉这道题目啦

$O(n)$遍历，过程中做$O(n^2)$的背包，就好了，$ans=dp_{1,Q}$
注意一些小细节

我记得神仙$kblack$讲过$O(n^2)$的做法，但是我忘了，只能$O(n^3)$了
Code：

#include <bits/stdc++.h>
#define maxn 110
using namespace std;
struct Edge{
	int to, next, len;
}edge[maxn << 1];
int num, head[maxn], n, m, size[maxn], dp[maxn][maxn], ans;

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 addedge(int x, int y, int z){ edge[++num] = (Edge){y, head[x], z}, head[x] = num; }

void dfs(int u, int pre){
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if (v != pre){
			dfs(v, u);
			size[u] += size[v] + 1;
			for (int j = min(m, size[u]); j; --j)
				for (int k = 0; k <= min(j - 1, size[v]); ++k)
					dp[u][j] = max(dp[u][j], dp[v][k] + dp[u][j - k - 1] + edge[i].len);
		}
	}
}

int main(){
	n = read(), m = read();
	for (int i = 1; i < n; ++i){
		int x = read(), y = read(), z = read();
		addedge(x, y, z), addedge(y, x, z);
	}
	dfs(1, 0);
	printf("%d\n", dp[1][m]);
	return 0;
}

来源：CSDN

作者：ModestCoder_

链接：https://blog.csdn.net/ModestCoder_/article/details/100807715