Round #201 (Div.2) B. Fixed Points

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本文介绍了一个算法问题，通过至多一次元素交换来最大化排列中的固定点数量。文章提供了完整的实现思路与C++代码示例，并解释了如何高效地解决这个问题。

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A permutation of length n is an integer sequence such that each integer from 0 to (n - 1) appears exactly once in it. For example, sequence [0, 2, 1] is a permutation of length 3 while both [0, 2, 2] and [1, 2, 3] are not.

A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We'll get a definition of a fixed point in a permutation. An integer i is a fixed point of permutation a₀, a₁, ..., a_n - 1 if and only if a_i = i. For example, permutation [0, 2, 1] has 1 fixed point and permutation [0, 1, 2] has 3 fixed points.

You are given permutation a. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in the resulting permutation. Note that you are allowed to make at most one swap operation.

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁵). The second line contains n integers a₀, a₁, ..., a_n - 1 — the given permutation.

Output

Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation.

Sample test(s)

Input

5
0 1 3 4 2

Output

————————————————————CF的分割线————————————————————

思路：交换的两个数仅一次，使得数组中和其下标相等的数字最多。两个for循环扫会超时，所以只能扫一次。另开一个vis数组，以和下标不一样的数字为下标储存这个数字的下标，即vis[a[i]] = i, 这样在扫一次的时候只要判断vis[i]是否等于a[i]，就可以知道交换这两个数字能不能增加两个fixed points。既然记录了错位的数字错在哪个下标，那么扫一次的时候直接利用这个下标去看数字就行了。

PS：还要讨论一下所有数字都是正确的情况。

代码如下：

/*
ID: j.sure.1
PROG: 
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <iostream>
using namespace std;
/****************************************/
const int N = 100005;
int a[N], vis[N];

int main()
{
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	int n, cnt = 0, ans = 0;
	bool ok = false;
	scanf("%d", &n);
	for(int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		if(a[i] != i) {
			ok = true;
			vis[a[i]] = i;
		}
		else
			ans++;
	}
	if(!ok)
		printf("%d\n", ans);
	else {
		bool flag = false;
		for(int i = 0; i < n; i++) {
			if(a[i] != i) {
				if(vis[i] == a[i]) {
					flag = true;
					break;
				}
			}
		}
		if(flag)
			printf("%d\n", ans+2);
		else
			printf("%d\n", ans+1);
	}
	return 0;
}