NOIP2015提高组.信息传递

题目

517. 信息传递

算法标签: 并查集, $Tarjan$算法, $scc$强连通分量

思路

使用强连通分量算法求环上点的数量, 对每个连通分量求最小点数

$Tarjan$算法求解代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>using namespace std;const int N = 200010, M = N;int n;
int head[N], ed[M], ne[M], idx;
int stk[N], top;
int low[N], dfn[N], timestamp;
int id[N], scc_cnt, cnt[N];
bool in_stk[N];void add(int u, int v) {ed[idx] = v, ne[idx] = head[u], head[u] = idx++;
}void tarjan(int u) {dfn[u] = low[u] = ++timestamp;stk[++top] = u, in_stk[u] = true;for (int i = head[u]; ~i; i = ne[i]) {int v = ed[i];if (!dfn[v]) {tarjan(v);low[u] = min(low[u], low[v]);}else if (in_stk[v]) low[u] = min(low[u], dfn[v]);}if (dfn[u] == low[u]) {++scc_cnt;int v;do {v = stk[top--];in_stk[v] = false;id[v] = scc_cnt;cnt[scc_cnt]++;}while (v != u);}
}int main() {ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);memset(head,-1, sizeof head);cin >> n;for (int i = 1; i <= n; ++i) {int t;cin >> t;add(i, t);}for (int i = 1; i <= n; ++i) {if (!dfn[i]) tarjan(i);}int ans = n;for (int i = 1; i <= scc_cnt; ++i) {if (cnt[i] > 1) ans = min(ans, cnt[i]);}cout << ans << "\n";return 0;
}