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+;
 int pos[M];

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

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

     return ;
 }