[HAOI2010]软件安装

题面：Luogu
题解：缩点+树形dp
题目还是很简单的，缩一下点，在新图上跑dp
这个dp连我都能一眼，可知难度多低
注意缩完点，建完新图后，再新建一个根，连到每一个入度为0的点上，再dfs

#include<bits/stdc++.h>
using namespace std;
inline void read(int& x)
{
	x = 0; char c = getchar();
	while (!isdigit(c)) c = getchar();
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}
#define maxn 2005
int n, m;
struct Graph
{
	struct Edge
	{
		int fr, to;
	}eg[maxn << 1];
	int head[maxn], edgenum;
	inline void add(int fr, int to)
	{
		eg[++edgenum] = { head[fr],to };
		head[fr] = edgenum;
	}
	int dfn[maxn], low[maxn], dfn_clock, col[maxn], colnum, vis[maxn];
	stack<int> sta;
#define to eg[i].to
	void paint(int rt)
	{
		vis[rt] = 0; sta.pop();
		col[rt] = colnum;
	}
	void Tarjan(int rt)
	{
		low[rt] = dfn[rt] = ++dfn_clock;
		vis[rt] = 1; sta.push(rt);
		for (int i = head[rt]; i; i = eg[i].fr)
		{
			if (!dfn[to]) Tarjan(to), low[rt] = min(low[rt], low[to]);
			else if (vis[to]) low[rt] = min(low[rt], dfn[to]);
		}
		if (low[rt] == dfn[rt])
		{
			++colnum;
			while (sta.top() != rt) paint(sta.top());
			paint(rt);
		}
	}
}G, g;
int w[maxn], v[maxn], d[maxn];
int siz[maxn], sum[maxn], deg[maxn], dp[maxn][maxn];
void dfs(int rt)
{
	for (int i = siz[rt]; i <= m; ++i) dp[rt][i] = sum[rt];
	for (int i = g.head[rt]; i; i = g.eg[i].fr)
	{
		dfs(g.to);
		for (int j = m; j >= siz[rt]; --j)
			for (int k = 0; k <= j - siz[rt]; ++k)
				dp[rt][j] = max(dp[rt][j], dp[rt][j - k] + dp[g.to][k]);
	}
}
int main()
{
	read(n), read(m);
	for (int i = 1; i <= n; ++i) read(w[i]);
	for (int i = 1; i <= n; ++i) read(v[i]);
	for (int i = 1; i <= n; ++i)
	{
		read(d[i]);
		if (!d[i]) continue;
		G.add(d[i], i);
	}
	for (int i = 1; i <= n; ++i)
		if (!G.dfn[i]) G.Tarjan(i);
	for (int i = 1; i <= n; ++i)
	{
		siz[G.col[i]] += w[i], sum[G.col[i]] += v[i];
		if (G.col[d[i]] == G.col[i] || !d[i]) continue;
		g.add(G.col[d[i]], G.col[i]); ++deg[G.col[i]];
	}
	for (int i = 1; i <= G.colnum; ++i)
		if (!deg[i]) g.add(G.colnum + 1, i);
	dfs(G.colnum + 1);
	printf("%d\n", dp[G.colnum + 1][m]);
	return 0;
}