数据结构 10-排序6 Sort with Swap(0, i) (25 分)

Given any permutation of the numbers {0, 1, 2,..., N−1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the following way:

Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4) => {0, 1, 2, 3, 4}

 

Now you are asked to find the minimum number of swaps need to sort the given permutation of the first N nonnegative integers.

Input Specification:

Each input file contains one test case, which gives a positive N (≤10​5​​) followed by a permutation sequence of {0, 1, ..., N−1}. All the numbers in a line are separated by a space.

Output Specification:

For each case, simply print in a line the minimum number of swaps need to sort the given permutation.

Sample Input:

10
3 5 7 2 6 4 9 0 8 1

 

Sample Output:

9

 

 

题目的目的是 交换次序的公式 

参考文章https://blog.csdn.net/qq_39339575/article/details/90343070

单元环 d 

多元环 k

如果多元环不含0(flag==true) 交换次数为n-d+k;

如果多元环含0 (flag==false) 交换次数为n-d+k-2;

 

 

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(){
    int n;
    int temp;
    bool flag{false};
    int d{0},k{0};
    cin >> n;
    vector<int> list(n);
    for(int i=0;i<n;i++){
        cin >> temp;
        list[i]=temp;
    }
    for(int i=0;i<n;i++){//单元环数量
        if(list[i]==i){
            d++;
            if(i==0)flag=true;
        }
    }
    for(int i=0;i<n;i++){//多元环的数量
        if(list[i]!=i){
            k++;
            while(list[i]!=i){
                int temp=list[i];
                list[i]=i;
                i=temp;
            }
        }
    }
    if(flag) cout << n-d+k<<endl;//无0的多元环
    else{//有0的多元环
        cout << n-d+k-2<<endl;
    }
    return 0;
}