【BZOJ5290】【HNOI2018】—道路(树形dp)

传送门

不说了吧,noipnoip难度dpdp
f[i][j][k]f[i][j][k]表示ii为根,两种路分别有j,kj,k个的最小值
大力dpdp就完了

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define int long long
inline int read(){
	char ch=getchar();
	int res=0,f=1;
	while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=getchar();
	return res*f;
}
const int N=20005;
const int M=45;
int n,f[N][M][M],a[N<<1],b[N<<1],c[N<<1],lc[N],rc[N];
inline int calc(int u,int A,int B){
	if(u>n)return c[u]*(a[u]+A)*(b[u]+B);
	return f[u][A][B];
}
void dfs(int u,int a,int b){
	if(u>=n)return;
	int v1,v2;
	dfs(v1=lc[u],a+1,b);
	dfs(v2=rc[u],a,b+1);
	for(int i=0;i<=a;i++)
		for(int j=0;j<=b;j++)
			f[u][i][j]=min(calc(v1,i+1,j)+calc(v2,i,j),calc(v1,i,j)+calc(v2,i,j+1));
}
signed main(){
	n=read();
	for(int i=1;i<n;i++){
		lc[i]=read(),rc[i]=read();
		if(lc[i]<0)lc[i]=-lc[i]+n;
		if(rc[i]<0)rc[i]=-rc[i]+n;
	}
	for(int i=1;i<=n;i++){
		a[i+n]=read(),b[i+n]=read(),c[i+n]=read();
	}
	dfs(1,0,0);
	cout<<f[1][0][0];
}
posted @ 2019-03-01 17:04  Stargazer_cykoi  阅读(136)  评论(0编辑  收藏  举报