7-16 Sort with Swap(0, i) (25分) 与零交换

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

能够发现序列是由几个环组成，每个环内的元素排好序就可以使得环内元素在正确的位置上，走一遍样例就可以发现，一个环内的元素相互交换最少需要元素数-1次操作就可以排好序，由于只能和0交换，如果这个环包括0，那就需要交换元素数-1次，否则的话，需要把0交换进环，再交换出去，需要交换元素数-1+2次即元素数+1。如果一个元素本来就在正确位置那就不用交换。
代码：

#include <stdio.h>
#include <stdlib.h>

int main() {
    int n,s[100000],ans = 0;
    scanf("%d",&n);
    for(int i = 0;i < n;i ++) {
        scanf("%d",&s[i]);
    }
    for(int i = 0;i < n;i ++) {
        int k = s[i];
        int c = 0;
        while(s[k] != k) {
            c ++;
            int t = k;
            k = s[k];
            s[t] = t;
        }
        if(c == 0) continue;
        if(i == 0) ans += c - 1;
        else ans += c + 1;
    }
    printf("%d",ans);
}