【JSOI2016】最佳团体 [分数规划 树形DP]

P4322 [JSOI2016]最佳团体
二分+树形dp

树形dp部分就和选课一样 就是r按照最保险的来赋的话会超时...

然后看了大佬的代码大佬r=3.3 QAQ

会T的那种

#define ll long long
#define Abs(x) ((x)<0?-(x):(x))
#define Max(x,y) ((x)>(y)?(x):(y))
#define Min(x,y) ((x)<(y)?(x):(y))
const int N=2500+5,M=1000+5,INF=1e9+7,inf=0x3f3f3f3f;
const double eps=1e-5;
int n,K,tj[N],a[N],b[N];
template void rd(t &x){
    x=0;int w=0;char ch=0;
    while(!isdigit(ch)) w|=ch=='-',ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    x=w?-x:x;
}
int head[N],tot=0;
struct edge{int v,nxt;}e[N<<1];
void add(int u,int v){
	e[++tot]=(edge){v,head[u]},head[u]=tot;
}
double d[N],f[N][N];
int sz[N];
void dfs(int u,int fa){
	sz[u]=1;
	for(int i=head[u],v;i;i=e[i].nxt){
		v=e[i].v;
		if(v==fa) continue;
		dfs(v,u);
		sz[u]+=sz[v];
		for(int j=Min(K+1,sz[u]);j;--j)
		for(int k=0;k0;
}
int main(){
	freopen("in.txt","r",stdin);
	rd(K),rd(n);
	double l=0.0,r=3.3,mid;
	for(int i=2;i<=n+1;++i) rd(b[i]),rd(a[i]),rd(tj[i]),add(++tj[i],i);//,,add(i,tj[i]);r=Max(r,(double)a[i]/b[i])
	while(r-l>=eps){
		mid=(l+r)/2;
		if(check(mid)) l=mid;
		else r=mid;
	}
	printf("%.3f",l);
	return 0;
}

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
#include<stack>
#include<algorithm>
using namespace std;
#define ll long long
#define Abs(x) ((x)<0?-(x):(x))
#define Max(x,y) ((x)>(y)?(x):(y))
#define Min(x,y) ((x)<(y)?(x):(y))
const int N=2500+5,M=1000+5,INF=1e9+7,inf=0x3f3f3f3f;
const double eps=1e-5;
int n,K,tj[N],a[N],b[N];
template <class t>void rd(t &x){
    x=0;int w=0;char ch=0;
    while(!isdigit(ch)) w|=ch=='-',ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    x=w?-x:x;
}

int head[N],tot=0;
struct edge{int v,nxt;}e[N<<1];
void add(int u,int v){
	e[++tot]=(edge){v,head[u]},head[u]=tot;
}

double d[N],f[N][N];
int sz[N];
void dfs(int u,int fa){
	sz[u]=1;
	for(int i=head[u],v;i;i=e[i].nxt){
		v=e[i].v;
		if(v==fa) continue;
		dfs(v,u);
		sz[u]+=sz[v];
		for(int j=Min(K+1,sz[u]);j;--j)
		for(int k=0;k<Min(sz[v]+1,j);++k)
		f[u][j]=Max(f[u][j],f[u][j-k]+f[v][k]);
	}
}

bool check(double mid){
	memset(f,0xc2,sizeof(f));
	for(int i=1;i<=n+1;++i) f[i][1]=d[i]=(double)a[i]-mid*b[i],f[i][0]=0.0;
	dfs(1,0);
	return f[1][K+1]>0;
}

int main(){
	freopen("in.txt","r",stdin);
	rd(K),rd(n);
	double l=0.0,r=3.3,mid;
	for(int i=2;i<=n+1;++i) rd(b[i]),rd(a[i]),rd(tj[i]),add(++tj[i],i);//,,add(i,tj[i]);r=Max(r,(double)a[i]/b[i])
	while(r-l>=eps){
		mid=(l+r)/2;
		if(check(mid)) l=mid;
		else r=mid;
	}
	printf("%.3f",l);
	return 0;
}

另一种先处理出时间戳的树形dp 先咕一下QAQ