PAT 1067 Sort with Swap(0, i)（25 分）（用指定方法排序）

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

题解：用题目给的指定方法排序，只用0与其它位交换，问最小的交换次数。定义一个变量left记录当前数字没有归位的数量。使交换次数最小的办法是让0与本来应该在它当前所在位的数字交换，若0在其本位，则往右寻找第一个不是在本位的数字交换。可以直接使用#include<algorithm>头文件下的swap()函数。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=100005;
int pos[maxn];
int main()
{
    int n,num;
    int ans=0;
    
    cin>>n;
    int left=n-1;  //除0以外未归位的个数
    for(int i=0;i<n;i++)
    {
        cin>>num;
        pos[num]=i;
        if(num==i&&num!=0)
            left--;  //未归位的数字个数
    }
    int k=1;         //除0以外当前不在本位最小的数
    while(left>0)
    {
        if(pos[0]==0) //如果0在本位
        {
            while(k<n)
            {
                if(pos[k]!=k)
                {
                    swap(pos[0],pos[k]);
                    ans++;
                    break;
                }
                k++;
            }
        }
        while(pos[0]!=0)
        {
            swap(pos[0],pos[pos[0]]);
            ans++;
            left--;
        }
    }
    cout<<ans<<endl;
    return 0;
}