BZOJ 2427: [HAOI2010]软件安装( dp )

软件构成了一些树和一些环, 对于环我们要不不选, 要么选整个环. 跑tarjan缩点后, 新建个root, 往每个入度为0的点(强连通分量) 连边, 然后跑树dp( 01背包 )

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#include<cstdio>

#include<cstring>

#include<stack>

#include<algorithm>

#include<vector>

#include<iostream>

  

#define rep(i, n) for(int i = 0; i < n; i++)

#define clr(x, c) memset(x, c, sizeof(x))

#define Rep(i, n) for(int i = 1; i <= n; ++i)

  

using namespace std;

 

const int maxn = 109, maxm = 509;

 

vector<int> G[maxn];

stack<int> S;

int dfn[maxn], low[maxn], scc[maxn], dfs_clock = 0, N = 0, n, m;

int W[maxn], V[maxn], w_t[maxn], v_t[maxn], d[maxn][maxm];

bool F[maxn];

 

void tarjan(int x) {

dfn[x] = low[x] = ++dfs_clock;

S.push(x);

for(vector<int>::iterator it = G[x].begin(); it != G[x].end(); it++)

if(!dfn[*it]) {

tarjan(*it);

low[x] = min(low[x], low[*it]);

} else if(!scc[*it])

low[x] = min(low[x], dfn[*it]);

if(low[x] == dfn[x]) {

N++;

int t;

do {

t = S.top(); S.pop();

scc[t] = N;

w_t[N] += W[t];

v_t[N] += V[t];

} while(t != x);

}

}

void TARJAN() {

clr(w_t, 0), clr(v_t, 0);

clr(scc, 0), clr(dfn, 0);

Rep(i, n) if(!dfn[i]) tarjan(i);

}

 

struct edge {

int to;

edge* next;

} E[maxn << 2], *pt = E, *head[maxn];

 

inline void add_edge(int u, int v) {

F[pt->to = v] = true;

pt->next = head[u];

head[u] = pt++;

}

 

void build() {

clr(F, 0), clr(head, 0);

Rep(i, n) 

   for(vector<int>::iterator it = G[i].begin(); it != G[i].end(); it++)

       if(scc[*it] != scc[i]) add_edge(scc[i], scc[*it]);

Rep(i, N) if(!F[i])

   add_edge(0, i);

}

 

void dp(int x) {

for(int i = v_t[x]; i <= m; i++) d[x][i] = w_t[x];

for(edge* e = head[x]; e; e = e->next) {

dp(e->to);

for(int h = m; h >= v_t[x]; h--)

   for(int t = 0; t <= h - v_t[x]; t++) 

   d[x][h] = max(d[x][h], d[x][h - t] + d[e->to][t]);

}

}

 

int main() {

// freopen("test.in", "r", stdin);

cin >> n >> m;

Rep(i, n) scanf("%d", V + i);

Rep(i, n) scanf("%d", W + i);

Rep(i, n) {

int v;

scanf("%d", &v);

if(v) G[v].push_back(i);

}

TARJAN();

build();

clr(d, 0), dp(0);

cout << *max_element(d[0], d[0] + m + 1) << "\n";

return 0;

}

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