tarjan

1
#include <bits/stdc++.h>
2
using namespace std;
3
const int N = 1e4 + 10, E = 2e5 + 10;
4
int n, m;
5
struct A {
6
  int first[N], to[E], nxt[E], cnt;
7
  void adde(int u, int v) {
8
    ++cnt;
9
    to[cnt] = v;
10
    nxt[cnt] = first[u];
11
    first[u] = cnt;
12
  }
13
} o, asd; int w[N];
14
queue<int> q;
15
int dfn[N], id, low[N], insta[N], sta[N], top;
66 collapsed lines
16
int col[N], colcnt, colw[N];
17
int f[N], in[N];
18
void dfs(int u) {
19
  dfn[u] = low[u] = ++id;
20
  sta[++top] = u; insta[u] = 1;
21
  for (int p = o.first[u]; p; p = o.nxt[p]) {
22
    int v = o.to[p];
23
    if (dfn[v] == 0) {
24
      dfs(v);
25
            // v 在搜索树上搜完了
26
            // low[u] 的定义：搜索树上 u 的子树最多经过一条后向边能到达的时间戳
27
      low[u] = min(low[u], low[v]);
28
    }
29
    else {
30
      if (insta[v])
31
        low[u] = min(low[u], dfn[v]);
32
            // 注意！这里 insta 不是搜索栈，而是未确定强连通块的点的栈
33
            // 为什么这里是 insta 才更新呢？
34
            // 如果不 insta[v]，注意此时 dfn[v] != 0 说明 v 已经访问过了，然而这说明访问 v 所在强连通块并不包含 u
35
            // 如果 insta[v]，就很抽象了，我也不好说，你自己记住吧
36
    }
37
  }
38
  if (low[u] == dfn[u]) {
39
    ++colcnt;
40
    do {
41
      col[sta[top]] = colcnt;
42
      colw[colcnt] += w[sta[top]];
43
      insta[sta[top]] = 0;
44
    } while (sta[top--] != u);
45
  }
46
}
47
int main() {
48
  ios::sync_with_stdio(0); cin >> n >> m;
49
  for (int i = 1; i <= n; ++i) {
50
    cin >> w[i];
51
  } for (int i = 1; i<= m; ++i) {
52
    int u, v; cin >> u >> v; o.adde(u, v);
53
  }
54
  for (int i = 1; i <= n; ++i) {
55
    if (not dfn[i])
56
      dfs(i);
57
  }
58
  for (int u = 1; u <= n; ++u) for (int p = o.first[u]; p; p = o.nxt[p])
59
  {
60
    int v = o.to[p];
61
    if (col[u] != col[v])
62
      asd.adde(col[u], col[v]), ++in[col[v]];
63
  }
64
  int ans = 0;
65
  for (int i = 1; i <= colcnt; ++i)
66
    if (in[i] == 0) q.push(i);
67
  while (not q.empty()) {
68
    int u = q.front(); q.pop();
69
    f[u] += colw[u];
70
    if (f[u] > ans) ans = f[u];
71
    for (int p = asd.first[u]; p ; p= asd.nxt[p]) {
72
      int v = asd.to[p];
73
      if (f[u] > f[v])
74
        f[v] = f[u];
75
      --in[v];
76
      if (in[v] == 0)  q.push(v);
77
    }
78
  }
79
  cout << ans << endl;
80
  return 0;
81
}