PAT A1067

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:
9

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
#include "stdio.h"
//#include "math.h"
#include "string.h"
#include "algorithm"
using namespace std;
const int maxn = 100010;
int pos[maxn];
int main(){
int n, ans = 0;
scanf("%d", &n);
int left = n - 1, num;//left存放除0以外不在本位的数
for (int i = 0 ; i < n; i++) {
scanf("%d", &num);
pos[num] = i;
if (num == i && num !=0) {
left--;
}
}
int k = 1;//k存放除了0意外当前不在本位上的最小的数
while (left > 0) {
//如果0在本位上,则寻找一个当前不在本位上的数与0交换
if (pos[0] == 0) {
while (k < n) {
if (pos[k] != k) {
swap(pos[0], pos[k]);
ans++;
break;
}
k++;
}
}
//只要0不在本位,就将0所在的位置的数的当前所处位置与0的位置交换
while (pos[0] != 0) {
swap(pos[0], pos[pos[0]]);
ans++;
left--;
}
}
printf("%d\n", ans);

return 0;
}