原题链接:忙碌的小L
hqz nb
#include <bits/stdc++.h>
constexpr auto Inf = 0X3F3F3F3F;
using namespace std;
typedef long long LL;
namespace IO {
inline LL read() {
LL o = 0, f = 1; char c = getchar();
while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
while (c > '/' && c < ':') { o = o * 10 + c - '0'; c = getchar(); }
return o * f;
}
inline char recd() {
char o; while ((o = getchar()) != 'G' && o != 'C'); return o;
}
}
using namespace IO;
const int SIZE = 1E5 + 7;
vector<int> G[SIZE];
int dfn[SIZE], scc[SIZE], low[SIZE], dfn_index, scc_index;
int on_stack[SIZE], top, stak[SIZE];
void tarjan(int u) {
on_stack[u] = 1, stak[++top] = u;
dfn[u] = low[u] = ++dfn_index;
for (int e = 0; e < G[u].size(); e++) {
int v = G[u][e];
if (!dfn[v])
tarjan(v), low[u] = min(low[u], low[v]);
else if (on_stack[v])
low[u] = min(low[u], dfn[v]);
}
if (low[u] == dfn[u]) {
++scc_index;
while (stak[top] != u)
on_stack[stak[top]] = 0, scc[stak[top--]] = scc_index;
on_stack[stak[top]] = 0, scc[stak[top--]] = scc_index;
}
}
int indegree[SIZE];
int main() {
int N = read();
for (int u = 1, v; u <= N; u++)
while (233) {
v = read();
if (!v) break;
G[u].push_back(v);
}
for (int u = 1; u <= N; u++)
if (!dfn[u]) tarjan(u);
for (int u = 1; u <= N; u++)
for (int e = 0; e < G[u].size(); e++) {
int v = G[u][e];
if (scc[u] != scc[v])
indegree[scc[v]]++;
}
int ans = 0;
for (int p = 1; p <= scc_index; p++)
ans += indegree[p] == 0;
printf("%d\n", ans);
}0.0分
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