B - Fixed Points

原创于 2017-03-16 10:13:08 发布 · 288 阅读

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本文介绍了一个算法问题，即通过至多一次元素交换操作来最大化排列中的固定点数量。文中提供了详细的解析过程及C++实现代码。

Description

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 Input

Input

5
0 1 3 4 2

Output

题意：

如果i=a[i]那么i就是一个固定点，最多交换一次位置，取得最多的固定点

分析：

将不是固定点的放在一起map里，找关键字，看是否符合

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,i,a[100005],s,p,t,o;
    map<int,int>v;
    map<int,int>::iterator it1,it2;
    while(cin>>n)
    {
        for(i=0,s=0,t=0,o=0,p=0;i<n;i++)
            {
                cin>>a[i];
                if(a[i]!=i)
                {
                    v.insert(make_pair(i,a[i]));
                    t=1;
                }
                else
                    s++;
            }
        while(!v.empty())
        {
            it2=v.begin();
            it1=v.find((*it2).second);
            if(it1!=v.end())
            {
                o=1;
                if((*it2).first==(*it1).second)
                    {
                        p=1;
                        break;
                    }
                else
                {
                    v.erase(it2);
                    v.erase(it1);
                }
            }
            else
                v.erase(it2);
        }
        v.clear();
        if(p)
            cout<<s+2<<endl;
        else
            if(o)
            cout<<s+1<<endl;
        else
            cout<<s<<endl;

    }
}

感受：

其实可以不用map，map还是费事，其实存数组了就好