1067 Sort with Swap(0, i) 贪心

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 Nnonnegative 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 n;
int find(int *arr,int s,int e)
{
    for(int i=s;i<e;i++)
    {
        if(arr[i]!=i)
        {
            return i;
        }
    }
    return 0;
}
void swap(int &a,int &b)
{
    int temp;
    temp=a;
    a=b;
    b=temp;
}
int main()
{
    int cnt=0,temp;
    cin>>n;
    int * a=new int[n];
    for(int i=0;i<n;i++)
    {
        cin>>temp;
        a[temp]=i;//因为这样查找temp时间复杂度最少
    }
    int pos=1;
    pos=find(a,1,n);
    while(pos)
    {
            if(a[0]==0)
            {
                a[0]=a[pos];
                a[pos]=0;
                cnt++;
            }
            while(a[0]!=0)
            {
                temp=a[0];
                a[0]=a[temp];
                a[temp]=temp;
                cnt++;
            }
            pos=find(a,pos,n);
    }
    cout<<cnt<<endl;
    return 0;
 } 

不知道怎么证明是cnt最小的