2020 cattle off winter training camp 6.B-- basis algorithm [Fig Tree & ring memory of dfs]

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Title Description

There are N points a directed graph, each edge point out only one
number you need to find a simple FIG longest path containing points
(1≤N≤1,000,000)

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Enter a description:

A number N of the first row
next N rows, each row a positive integer, the i + 1 row numbers i-th point represents the number of end edges
(starting from a reference point)

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Output Description:

A digit line, the path length of the longest simple

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Entry

3
2
3
2

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Export

3

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answer

• First, it is a tree ring
• Each point of the graph are to have a ring within the tree forest 1
• From a starting point, keep going and will certainly come to a ring
• When we arrived at the ring when statistics can answer on a ring
• The use of memory dfs
• For details, see Notes
  

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AC-Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const int maxn = 1e6 + 7;
const int mod = 1e9 + 7;

int to[maxn];
int vis[maxn];
int ans[maxn];
int dfs(int now) {
	vis[now] = 1;
	if (vis[to[now]]) { // 下一个点 走过
		if (ans[to[now]]) { // 同一条路径先算子节点。如果>1说明不是一条路径
			ans[now] = ans[to[now]] + 1;
		}
		else { // 走到环
			int t = now, num = 1;
			while (to[t] != now) { // 走一圈环，记录环的长度
				t = to[t];
				++num;
			}
			ans[now] = num;
			t = now;
			while (to[t] != now) { // 再走一遍环，更新环上点的答案
				t = to[t];
				ans[t] = num;
			}
		}
	}
	else {
		dfs(to[now]);
		if (!ans[now]) ans[now] = ans[to[now]] + 1;
	}
	return ans[now];
}
int main() {
	ios;
	int n;	while (cin >> n) {
		memset(vis, 0, sizeof vis);
		memset(ans, 0, sizeof ans);
		for (int i = 1; i <= n; ++i)		cin >> to[i];
		int ans = 0;
		for (int i = 1; i <= n; ++i) {
			if (!vis[i])	ans = max(ans, dfs(i));
		}
		cout << ans << endl;
	}
}