思路是tarjan缩点然后dp，应该不需要拓扑排序吧。。。

#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, b, a) for (int i = b; i >= a; i--)
using namespace std;
using pii = pair<int, int>;
using graph = vector<vector<pii>>;
using ugraph = vector<vector<int>>;
using ll = long long;
constexpr int MAXN = 2E5 + 5;
constexpr ll INF = 1E18;

int tim, tot;  // 时间戳, scc个数
stack<int> st;
bool in_stack[MAXN];
int dfn[MAXN], low[MAXN], num[MAXN], color[MAXN], sum[MAXN];
void tarjan(int x, ugraph &g) {
    st.push(x), in_stack[x] = true;
    dfn[x] = low[x] = ++tim;
    for (auto y : g[x]) {
        if (dfn[y] == 0) {
            tarjan(y, g);
            low[x] = min(low[x], low[y]);
        } else if (in_stack[y]) {
            low[x] = min(low[x], low[y]);
        }
    }
    if (dfn[x] == low[x]) {
        tot++;
        while (st.top() != x) {
            sum[tot] += num[st.top()];
            color[st.top()] = tot;
            st.pop(), in_stack[x] = false;
        }
        sum[tot] += num[x];
        color[x] = tot;
        st.pop(), in_stack[x] = false;
    }
}
int dp[MAXN];
void dfs(int x, ugraph &g) {
    dp[x] = sum[x];
    int res = 0;
    for (int y : g[x]) {
        if (!dp[y]) dfs(y, g);
        res = max(res, dp[y]);
    }
    dp[x] += res;
}
int u[MAXN], v[MAXN];
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    ugraph g(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> num[i];
    }
    for (int i = 1; i <= m; i++) {
        cin >> u[i] >> v[i];
        g[u[i]].emplace_back(v[i]);
    }
    for (int i = 1; i <= n; i++) {
        if (!dfn[i]) tarjan(i, g);
    }
    ugraph h(n + 1);
    for (int i = 1; i <= m; i++) {
        if (color[u[i]] != color[v[i]]) {
            h[color[u[i]]].emplace_back(color[v[i]]);
        }
    }
    for (int i = 1; i <= tot; i++) {
        if (!dp[i]) dfs(i, h);
    }
    int ans = 0;
    for (int i = 1; i <= tot; i++) {
        ans = max(ans, dp[i]);
    }
    cout << ans;
    return 0;
}