题目

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分析

那么考虑二分答案，那么每个也就变成选择一群合法的招募人，使其和不少于0
然而$O(n^3)$O(n3)会超时
所以需要维护dfs序优化，时间复杂度$O(n^2)$O(n2)

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代码

#include <cstdio>
#include <cctype>
#include <cstring>
#define rr register
using namespace std;
const int N=2511;
struct node{int y,next;}e[N]; double p[N],dp[N][N];
int w[N],ls[N],c[N],dfn[N],ed[N],n,m,k,tot;
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
inline void add(int x,int y){
	e[++k]=(node){y,ls[x]},ls[x]=k;
}
inline void dfs(int x){
	dfn[x]=tot++;
	for (rr int i=ls[x];~i;i=e[i].next)
	    dfs(e[i].y);
	ed[dfn[x]]=tot;
}
inline bool check(double mid){
	for (rr int i=1;i<=n;++i) p[dfn[i]]=c[i]-mid*w[i];
	for (rr int i=1;i<=n+1;++i)
	for (rr int j=0;j<=m+1;++j) dp[i][j]=-1e7;
	for (rr int i=0;i<=n;++i)
	for (rr int j=0;j<=m+1&&j<=i;++j){
		if (dp[i][j]+p[i]>dp[i+1][j+1]) dp[i+1][j+1]=dp[i][j]+p[i];
		if (dp[i][j]>dp[ed[i]][j]) dp[ed[i]][j]=dp[i][j];
	}
	return dp[n+1][m+1]>1e-5;
}
signed main(){
	m=iut(); n=iut();
	memset(ls,-1,sizeof(ls));
	for (rr int i=1;i<=n;++i)
		w[i]=iut(),c[i]=iut(),add(iut(),i);
	dfs(0);
	rr double l=0,r=10000;
	while (l+1e-5<r){
		rr double mid=(l+r)/2;
		if (check(mid)) l=mid;
		    else r=mid;
	}
	return !printf("%.3lf",l);
}