[BZOJ2427]软件安装

Problem

每个软件都要安装某些软件才能安装，而且都有体积和价值，求安装的价值最大值

Solution

对于每个环，我们可以知道必须全部一起取或者不取，因此我们先用Tarjan缩点

然后我们用一个树形DP就可以解决了

Notice

注意这颗树是如果一个节点没取，后面就都不能取了

Code

#include<cmath>
#include<stack>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
#define travel2(i, u) for (reg i = head2[u]; i; i = edge2[i].next)
const int INF = 1e9;
const double eps = 1e-6, phi = acos(-1);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
int f[105][1005];
int Time = 0, num1 = 0, num2 = 0, scc = 0, n, m;
stack<int> Stack;
struct node
{
	int vet, next;
}edge[505], edge2[505];
int head[105], head2[105], dfn[105], low[105], in[105], belong[105], v[105], w[105], V[105], W[105], flag[105];
void addedge(int u, int v)
{
	edge[++num1].vet = v;
	edge[num1].next = head[u];
	head[u] = num1;
}
void add(int u, int v)
{
	in[v]++;
	edge2[++num2].vet = v;
	edge2[num2].next = head2[u];
	head2[u] = num2;
}
void Tarjan(int u)
{
	dfn[u] = low[u] = ++Time;
	Stack.push(u);
	flag[u] = 1;
	travel(i, u)
	{
		int v = edge[i].vet;
		if (!dfn[v])
		{
			Tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else if (flag[v]) low[u] = min(low[u], dfn[v]);
	}
	if (low[u] == dfn[u])
	{
		int t; scc++;
		do
		{
			t = Stack.top();
			Stack.pop();
			belong[t] = scc;
			flag[t] = 0;
			V[scc] += v[t];
			W[scc] += w[t];
		}while (t != u);
	}
}
void dp(int u)
{
	travel2(i, u)
	{
		int v = edge2[i].vet;
		dp(v);
		per(j, m - V[u], 0)
			rep(k, 0, j) f[u][j] = max(f[u][j], f[u][k] + f[v][j - k]);
	}
	per(j, m, V[u]) f[u][j] = f[u][j - V[u]] + W[u];
	per(j, V[u] - 1, 0) f[u][j] = 0;
}
int sqz()
{
	n = read(), m = read();
	rep(i, 1, n) v[i] = read();
	rep(i, 1, n) w[i] = read();
	rep(i, 1, n) addedge(read(), i);
	rep(i, 0, n)
		if (!dfn[i]) Tarjan(i);
	rep(u, 1, n)
		travel(i, u)
		{
			int v = edge[i].vet;
			if (belong[u] != belong[v]) add(belong[u], belong[v]);
		}
	rep(i, 1, scc) if (!in[i]) add(0, i);
	dp(0);
	printf("%d\n", f[0][m]);
	return 0;
}