10-排序6 Sort with Swap(0, i) (25 分)

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

Note

1. 精简版表排序，本质上是尽求移动次数最少，因此适合本题的swap
2. 交换的次数求解
   
3. 注意讨论a[0] = 0 , 若a[0] == 0 环的第二中情况就不存在了，因此交换的次数为 ni + 1 从1到k求和即等于所有多元环中元素的个数 + k，而其中多元环中元素的个数等于num - 单元环的个数 所以公式变为N - S + K

Code

#include<iostream>
using namespace std;
#define MAX 100100
int a[MAX];
int main() {
    int i,num, S = 0, K = 0, flag = 0;
    cin >> num;
    for(i = 0; i < num; i++){
        cin >> a[i];
        if(a[i] == i) {
            if(i == 0) flag = 1;
            S++;								//单元环个数
        }
    }
    for(i = 0; i < num; i++){
        if(a[i] != i){
            K++;								//多元环个数
            while(a[i] != i){
                int temp = a[i];
                a[i] = i;
                i = temp;
            }
        }
    }
    if(flag) cout << num - S + K;
    else cout << num - S + K - 2;
}