Codeforces - Captain Flint and Treasure

题目链接:Codeforces - Captain Flint and Treasure


如果不是输出路径的话,显然就是DAG上面dp。现在需要输出路径,我们就要知道选的先后顺序。

然后每个点就看之前的点贡献的正负,确定先后顺序,然后又是一个拓扑排序了。


AC代码:

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+10;
int n,a[N],b[N],deg[N],dp[N],res,d[N];
vector<int> g[N],g1[N];
void Top(){
	queue<int> q; for(int i=1;i<=n;i++) if(!deg[i]) q.push(i);
	while(q.size()){
		int u=q.front();	q.pop();	res+=a[u];
		for(int to:g[u]){
			if(--deg[to]==0)	q.push(to);
			a[to]+=max(0LL,a[u]);
			if(a[u]>=0)	g1[u].push_back(to),d[to]++;
			else	g1[to].push_back(u),d[u]++;
		}
	}
	cout<<res<<endl;
	for(int i=1;i<=n;i++)	if(!d[i])	q.push(i);
	while(q.size()){
		int u=q.front();	q.pop();	printf("%d ",u);
		for(int to:g1[u])	if(--d[to]==0)	q.push(to);
	}
}
signed main(){
	cin>>n;
	for(int i=1;i<=n;i++)	cin>>a[i];
	for(int i=1;i<=n;i++){
		cin>>b[i];
		if(b[i]!=-1)	g[i].push_back(b[i]),deg[b[i]]++;
	}
	Top();
	return 0;
}

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转载自blog.csdn.net/weixin_43826249/article/details/107723384
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