【题解】LuoGu4322：[JSOI2016]最佳团体

原题传送门
应该是一种叫$01分数规划$的玩意
二分答案
对于每个$mid$，要使$\frac{P}{S}>=mid$，就是$P-S\ast mid>=0$
对于每一个物品$i$，根据$p_i,s_i,mid可求出它的“权值”w_i=p_i-s_i\ast mid$
问题转化为，树上的DP，拿大小为$m+1$的联通块（把根节点0算进了）是的权值和最大
再看看这个最大的权值和是否$>=0$

设计状态：$dp[u][x]$表示以u为根的子树，取x个的权值和最大值
记一个size，转移不是很好讲，具体可以见代码
边界考虑一下：$dp[u][0]=0,dp[u][1]=w[u]$

Code：

#include <bits/stdc++.h>
#define maxn 2510
#define eps 1e-5
using namespace std;
double dp[maxn][maxn], w[maxn];
int size[maxn], s[maxn], p[maxn], num, head[maxn], n, m, fa;
struct Edge{
	int to, next;
}edge[maxn << 1];

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

void dfs(int u){
	size[u] = 1, dp[u][1] = w[u];
	for (int i = head[u]; i; i = edge[i].next){
		int v = edge[i].to;
		dfs(v);
		for (int j = size[u]; j; --j)//枚举不在v子树中取点个数
			for (int k = 0; k <= size[v]; ++k)//枚举在v子树中取点个数
				dp[u][j + k] = max(dp[u][j + k], dp[u][j] + dp[v][k]);
		size[u] += size[v];//size统计
	}
}

bool check(double mid){
	memset(dp, -10, sizeof(dp));
	for (int i = 0; i <= n; ++i) dp[i][0] = 0;
	for (int i = 1; i <= n; ++i) w[i] = 1.0 * p[i] - s[i] * mid;
	dfs(0);
	return dp[0][m] >= 0;
}

int main(){
	m = read(), n = read(); ++m;
	for (int i = 1; i <= n; ++i) s[i] = read(), p[i] = read(), fa = read(), addedge(fa, i);
	double l = 0, r = 1e4, ans = 0;
	while (l + eps <= r){
		double mid = (l + r) / 2;
		if (check(mid)) ans = mid, l = mid + eps; else r = mid - eps;
	}
	printf("%.3f\n", ans);
	return 0;
}