洛谷上这道题的空间限制是 125MB，但是在 DMOJ 上同一道题的空间限制只有 64M ，做法是tarjan缩点，萌新不会什么卡空间技巧，求助：（

#include <bits/stdc++.h>
using  namespace std;

const int N = 500001;
#define ll long long

int n, m;
vector<int> g[N];
int val[N], low[N], dfn[N], cnt, idx, bar[N], ins[N], bel[N];
int dp[N];
stack<int> s;

void dfs(int u) {
    dfn[u] = low[u] = ++idx;
    s.push(u);
    ins[u] = 1;
    for (auto v: g[u]) {
        if (!dfn[v]) dfs(v);
        if (ins[v]) low[u] = min(low[u], low[v]);
    }
    if (low[u] == dfn[u]) {
        ++cnt;
        int sval = 0;
        bool sbar = false;
        dp[cnt] = -(1 << 30);
        while (true) {
            int v = s.top();
            ins[v] = 0;
            bel[v] = cnt;
            sval += (ll)val[v];
            sbar |= bar[v];
            for (auto w: g[v]) if (bel[w] != 0 && bel[w] != cnt) {
                dp[cnt] = max(dp[cnt], dp[bel[w]]);
            }
            s.pop();
            if (v == u) break;
        }
        if (sbar) dp[cnt] = max(0, dp[cnt]);
        dp[cnt] += sval;
    }
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int u, v; cin >> u >> v;
        g[u].push_back(v);
    }
    for (int i = 1; i <= n; i++) cin >> val[i];
    int s, p;
    cin >> s >> p;
    for (int i = 1; i <= p; i++) {
        int x; cin >> x;
        bar[x] = 1;
    }
    dfs(s);
    cout << dp[bel[s]] << endl;
    return 0;
}