PAT_A1067#Sort with Swap(0, i)

Source:

PAT A1067 Sort with Swap(0, i) (25 分)

Description:

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

Keys:

• 贪心

Code:

/*
Data: 2019-07-22 19:54:12
Problem: PAT_A1067#Sort with Swap(0, i)
AC: 29:14

题目大意：
排序，只能交换0和任意其他数，给出需要的最少步数

基本思路：
1.0在几号位就跟相应的数字换位置，每次操作可以将一个数字归位；
2.若0回到了0号位，则再与任意一个不在自己位置上的数字交换；
3.重复第一步，直至所有数字都归位；
*/

#include<cstdio>
#include<algorithm>
using namespace std;
const int M=1e5+10;
int pos[M];

int main()
{
#ifdef    ONLINE_JUDGE
#else
    freopen("Test.txt", "r", stdin);
#endif

    int n,num,left=0,cnt=0,st=0;
    scanf("%d", &n);
    for(int i=0; i<n; i++)
    {
        scanf("%d", &num);
        pos[num]=i;
        if(num!=i && num!=0)
            left++;
    }
    while(left!=0)
    {
        if(pos[0]==0){
            for(int i=st; i<n; i++){
                if(pos[i]!=i){
                    swap(pos[0],pos[i]);
                    st=i;break;
                }
            }
            cnt++;
        }
        swap(pos[0],pos[pos[0]]);
        cnt++;
        left--;
    }
    printf("%d", cnt);

    return 0;
}