bzoj 3875 骑士游戏

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SPFA.

设第 $i$ 只怪物的消耗为 $f(i)$ ,则
$$
f(i)=\min\lbrace K_i,S_i+\sum_{j\in son(i)} f(j) \rbrace
$$
这个转移可能成环,存在后效性,用 SPFA 去转移,一个点更新之后,要把所有连向它的点入队更新答案.

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//%std
#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long ll;
inline ll read()
{
	ll out=0,fh=1;
	char jp=getchar();
	while ((jp>'9'||jp<'0')&&jp!='-')
		jp=getchar();
	if (jp=='-')
		fh=-1,jp=getchar();
	while (jp>='0'&&jp<='9')
		out=out*10+jp-'0',jp=getchar();
	return out*fh;
}
const int MAXN=2e5+10;
int n,in[MAXN];
ll dis[MAXN],s[MAXN];
vector<int> sons[MAXN],fa[MAXN];
queue<int> q;
void SPFA()
{
	for(int i=1;i<=n;++i)
	{
		q.push(i);
		in[i]=1;
	}
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		in[u]=false;
		ll tmp=s[u];
		for(int i=0;i<(signed)(sons[u].size());++i)
			tmp+=dis[sons[u][i]];
		if(tmp<dis[u])
		{
			dis[u]=tmp;
			for(int i=0;i<(signed)(fa[u].size());++i)
			{
				int v=fa[u][i];
				if(!in[v])
				{
					q.push(v);
					in[v]=1;
				}	
			}			
		}			
	}
}
int main()
{
	n=read();
	for(int i=1;i<=n;++i)
	{
		s[i]=read();
		dis[i]=read();
		int m=read();
		for(int k=0;k<m;++k)
		{
			int j=read();
			sons[i].push_back(j);
			fa[j].push_back(i);
		}
	}
	SPFA();
	cout<<dis[1]<<endl;
	return 0;
}