Atcoder Beginner Contest 341F Breakdown

problem link

The condition of choosing a subset of neighbors to pass down the piece is highly similar to a backpack problem, where each neighbor has a value contribution and a backpack weight.

Let us denote $f[i]$ as the maximum number of operations possible when one piece is put on node $i$. Then, $f[i]$ can be acquired by performing a 01 backpack process with its neighbors, with their $w$ value as weight and $f$ value as value.

This calculation can be completed in $\mathcal{O}(W_{max}\cdot m)$ time if we start from the nodes with smallest $w$ values. The final answer would be $\sum f[i]\cdot a[i]$

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<map>
using namespace std;
const long long Maxn=5e3+10;
struct node{
	long long pos,val;
	bool operator <(const node &x)const
	{
		return x.val>val;
	}
}s[Maxn];
vector <long long> e[Maxn];
long long a[Maxn],w[Maxn];
long long f[Maxn],g[Maxn];
long long n,m,ans;
inline long long read()
{
	long long s=0,w=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0' && ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*w;
}
int main()
{
	// freopen("in.txt","r",stdin);
	n=read(),m=read();
	while(m--)
	{
		long long x=read(),y=read();
		e[x].push_back(y);
		e[y].push_back(x);
	}
	for(long long i=1;i<=n;++i)
	{
		w[i]=read();
		s[i]=node{i,w[i]};
	}
	for(long long i=1;i<=n;++i)
	a[i]=read();
	sort(s+1,s+1+n);
	for(long long k=1;k<=n;++k)
	{
		long long x=s[k].pos;
		memset(g,0,sizeof(g));
		for(long long i=0;i<e[x].size();++i)
		{
			long long y=e[x][i];
			if(w[y]>=w[x])continue;
			for(long long j=w[x];j>=w[y];--j)
			g[j]=max(g[j],g[j-w[y]]+f[y]);
		}
		for(long long i=1;i<w[x];++i)
		f[x]=max(f[x],g[i]);
		++f[x];
		ans+=f[x]*a[x];
	}
	// for(long long i=1;i<=n;++i)
	// printf("%d ",f[i]);
	// putchar('\n');
	printf("%lld\n",ans);
	return 0;
}