Java(javac 1.8) 解法, 执行用时: 200ms, 内存消耗: 15036K, 提交时间: 2020-05-30 10:45:22

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			int N[] = new int[sc.nextInt()], count = 0;
			for (int i = 0; i < N.length; i++) {
				N[i] = sc.nextInt();
			}
			for (int i = 0; i < N.length; i++) {
				boolean[] visited = new boolean[N.length];
				int j = N[i];
				while (true) {
					if (j == 0) {
						count++;
						break;
					} else if (visited[j - 1]) {
						break;
					}
					visited[j - 1] = true;
					j = N[j - 1];
				}
			}
			System.out.println(count);
		}
	}
}

C++11(clang++ 3.9) 解法, 执行用时: 4ms, 内存消耗: 608K, 提交时间: 2020-06-12 11:54:34

#include <iostream>
#include <vector>
#include <cstdio>
using namespace std;
const int N = 1005;
vector<int> g[N];
int n;
bool vis[N];
int cnt;
void dfs(int x) {
	//printf("x = %d\n", x);
	for (int i = 0; i < g[x].size(); i++) {
		int y = g[x][i];
		if (vis[y] == false) {
			cnt++;
			vis[y] = true;
			dfs(y);
		}
	}
}
int main() {
	//freopen("Message Relay.txt", "r", stdin);
	scanf("%d", &n);
	int t;
	for (int i = 1; i <= n; i++) {
		scanf("%d", &t);
		g[t].push_back(i);
	}
	dfs(0);
	printf("%d\n", cnt);
	return 0;
}

C(clang 3.9) 解法, 执行用时: 3ms, 内存消耗: 380K, 提交时间: 2020-05-30 09:48:11

#include <stdio.h>
#define maxn 2010

int opt[maxn],tf[maxn],a[maxn],n,ans = 0;
int f(int p){
	if (opt[p] > 0)
	      return opt[p];
	if (a[p] == 0){
		tf[p] = 1;
		return opt[p] = 1;
	}
	if (tf[p] == 1){
		return opt[p] = 2;
	}
	tf[p] = 1;
	return opt[p] = f(a[p]);
}
int main(){
	scanf("%d",&n);
	for (int i = 1; i <= n; ++i)
	   scanf("%d",&a[i]);
	for (int i = 1; i <= n; ++i)
	   if (f(i) == 1){
	   	      ans++;
	   }
	printf("%d\n", ans);
	return 0;
}
 

C++14(g++5.4) 解法, 执行用时: 7ms, 内存消耗: 604K, 提交时间: 2020-06-04 13:28:55

#include<iostream>
#include<cstdio>
using namespace std;
int f[1005],mk[1005],vis[1005],v[1005];
int main(){
	int i,j,k,m,n,ans=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&f[i]);
	}
	for(i=1;i<=n;i++){
		v[i]=1;m=i;
		while(v[f[m]]==0&&f[m]!=0){
			v[m]=1;
			m=f[m];
		}
		if(f[m]!=0){
			for(j=1;j<=n;j++) if(v[j]) mk[j]=1;
		}
		for(j=1;j<=n;j++) v[j]=0;
	}
	for(i=1;i<=n;i++) if(mk[i]==0) ans++;
	printf("%d",ans);
	return 0;
}