传送门
题解
Tarjan把环缩成点，然后跑树形背包即可。
我用的树形背包是DFS序上搞的那种。
要注意dp数组初始化成-INF！
要注意dp顺推的时候也不要忘记看数组是否越界！
长太息以掩涕兮，
哀bug之难De...
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long ll;
#define enter putchar('
')
#define space putchar(' ')
template <class T>
void read(T &x){
char c;
bool op = 0;
while(c = getchar(), c > '9' || c < '0')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}
const int N = 105, M = 505, _INF = 0xc0c0c0c0;
int n, m, ans, _c[N], _w[N], c[N], w[N], fa[N], _adj[N], _nxt[N];
int dfn[N], low[N], idx, stk[N], top, bel[N], scc, cnt[N];
int adj[N], nxt[N], dp[N][M], sze[N], seq[N], tot;
bool ins[N];
void tarjan(int u){
dfn[u] = low[u] = ++idx;
stk[++top] = u, ins[u] = 1;
for(int v = _adj[u]; v; v = _nxt[v])
if(!dfn[v])
tarjan(v), low[u] = min(low[u], low[v]);
else if(ins[v])
low[u] = min(low[u], dfn[v]);
if(low[u] == dfn[u]){
++scc;
int v = 0;
while(v != u){
ins[v = stk[top--]] = 0;
bel[v] = scc, cnt[scc]++;
c[scc] += _c[v], w[scc] += _w[v];
}
}
}
void rebuild(){
static bool vis[N] = {0};
for(int i = 1; i <= n; i++)
if(cnt[bel[i]] > 1 && !vis[bel[i]])
nxt[bel[i]] = adj[0], adj[0] = bel[i], vis[bel[i]] = 1;
else if(cnt[bel[i]] == 1)
nxt[bel[i]] = adj[bel[fa[i]]], adj[bel[fa[i]]] = bel[i];
}
void dfs(int u){
seq[++tot] = u, sze[u] = 1;
for(int v = adj[u]; v; v = nxt[v])
dfs(v), sze[u] += sze[v];
}
int main(){
read(n), read(m);
for(int i = 1; i <= n; i++) read(_c[i]);
for(int i = 1; i <= n; i++) read(_w[i]);
for(int i = 1; i <= n; i++)
read(fa[i]), _nxt[i] = _adj[fa[i]], _adj[fa[i]] = i;
for(int i = 1; i <= n; i++)
if(!dfn[i]) tarjan(i);
rebuild();
dfs(0);
memset(dp, _INF, sizeof(dp));
dp[1][0] = 0;
for(int i = 1; i <= tot; i++){
for(int j = 0; j <= m; j++){
dp[i + sze[seq[i]]][j] = max(dp[i + sze[seq[i]]][j], dp[i][j]);
if(j + c[seq[i]] <= m)
dp[i + 1][j + c[seq[i]]] = max(dp[i + 1][j + c[seq[i]]], dp[i][j] + w[seq[i]]);
}
}
for(int j = 0; j <= m; j++)
ans = max(ans, dp[tot + 1][j]);
write(ans), enter;
return 0;
}