1067. Sort with Swap(0,*) (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (<=10⁵) 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:

 1 #include <stdio.h>
 2 #include<algorithm>
 3 using namespace std;
 4 int main()
 5 {
 6     int n;
 7     int loc[100000];
 8     while(scanf("%d",&n)!=EOF)
 9     {
10         
11 
12         int tem,i,left;
13         left = n-1;
14         for(i=0;i<n;i++)
15         {
16             scanf("%d",&tem);
17             loc[tem] = i;
18             if(tem == i && i!=0)
19                 --left;
20         }
21 
22         int count = 0;
23         int k = 1;
24         while(left > 0)
25         {
26             if(loc[0]==0)
27             {
28                while(k < n)
29                {
30                   if( loc[k] != k )
31                   {
32                       swap(loc[0],loc[k]);
33                       ++count;
34                       break;
35                   }
36                   ++k;
37                }
38             }
39 
40 
41             while(loc[0]!=0)
42             {
43                 swap(loc[0],loc[loc[0]]);
44                 ++count;
45                 --left;
46             }
47 
48         }
49 
50         printf("%d
",count);
51     }
52     return 0;
53 }