1067 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 (≤) 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、

#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
    int n; int ans=0;//最后的答案
    scanf("%d",&n);
    int rest =n-1;//还有几个数不在本位上

    int pos[n];
    for(int i=0;i<n;i++)
    {
        scanf("%d",&pos[i]);//数字i存放在pos[i]
        if(pos[i]==i)
            rest--;
    }

    int k =1;//k存放除了0之外的不在本位上的最小数字
    while(rest)
    {
        if(pos[0]==0)//如果0在本位上
        {//则寻找一个不在本位上的数和0交换
            while(k<n)
            {
                if(pos[k]!=k)//如果k不在本位上就和0交换
                {
                    swap(pos[0],pos[k]);
                    ans++;
                    break;
                }
                k++;
            }
       }

         while(pos[0]!=0)
            {
                swap(pos[0],pos[pos[0]]);
                ans++;
                rest--;
            }
        }


    printf("%d",ans);

    return 0;
}

这段代码只有22分，有2个点运行超时！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！