原題連結

在一個強連通分量裡的\(ATM\)機顯然都可被搶，所以先用\(tarjan\)找強連通分量並縮點，在縮點的後的\(DAG\)上跑最長路，然後掃一遍酒吧記錄答案即可。

#include<cstdio>
using namespace std;
const int N = 5e5 + 10;
struct eg{
    int x, y;
};
eg a[N];
int fi[N], di[N], ne[N], cfi[N], cdi[N], cne[N], va[N], sta[N], dis[N], dfn[N], low[N], bl[N], bar[N], ru[N], q[N], su[N], l, lc, ti, tp, st, SCC;
bool v[N];
inline int re()
{
    int x = 0;
    char c = getchar();
    bool p = 0;
    for (; c < '0' || c > '9'; c = getchar())
        p |= c == '-';
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return p ? -x : x;
}
inline void add(int x, int y)
{
    di[++l] = y;
    ne[l] = fi[x];
    fi[x] = l;
}
inline void add_c(int x, int y)
{
    cdi[++lc] = y;
    cne[lc] = cfi[x];
    cfi[x] = lc;
}
inline int minn(int x, int y)
{
    return x < y ? x : y;
}
inline int maxn(int x, int y)
{
    return x > y ? x : y;
}
void tarjan(int x)
{
    int i, y;
    dfn[x] = low[x] = ++ti;
    sta[++tp] = x;
    v[x] = 1;
    for (i = fi[x]; i; i = ne[i])
        if (!dfn[y = di[i]])
        {
            tarjan(y);
            low[x] = minn(low[x], low[y]);
        }
        else
            if (v[y])
                low[x] = minn(low[x], dfn[y]);
    if (!(dfn[x] ^ low[x]))
    {
        SCC++;
        do
        {
            y = sta[tp--];
            bl[y] = SCC;
            su[SCC] += va[y];
            v[y] = 0;
        } while (x ^ y);
    }
}
void spfa()
{
    int i, x, y, head = 0, tail = 1;
    q[1] = bl[st];
    dis[bl[st]] = su[bl[st]];
    while (head ^ tail)
    {
        x = q[++head];
        v[x] = 0;
        for (i = cfi[x]; i; i = cne[i])
        {
            y = cdi[i];
            if (dis[y] < dis[x] + su[y])
            {
                dis[y] = dis[x] + su[y];
                if (!v[y])
                {
                    q[++tail] = y;
                    v[y] = 1;
                }
            }
        }
    }
}
int main()
{
    int i, k, n, m, x, y, ma = 0;
    n = re();
    m = re();
    for (i = 1; i <= m; i++)
    {
        a[i].x = re();
        a[i].y = re();
        add(a[i].x, a[i].y);
    }
    for (i = 1; i <= n; i++)
        va[i] = re();
    st = re();
    k = re();
    for (i = 1; i <= k; i++)
        bar[i] = re();
    for (i = 1; i <= n; i++)
        if (!dfn[i])
            tarjan(i);
    for (i = 1; i <= m; i++)
    {
        x = bl[a[i].x];
        y = bl[a[i].y];
        if (x ^ y)
        {
            add_c(x, y);
            ru[y]++;
        }
    }
    spfa();
    for (i = 1; i <= k; i++)
        ma = maxn(ma, dis[bl[bar[i]]]);
    printf("%d", ma);
    return 0;
}