PAT A1067. Sort with Swap(0,*) (25)

题目链接

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

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100010;
int mhash[maxn];
int main() {
    int n, cnt = 0, m, count = 0;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &m);
        mhash[m] = i;
        if (m != i&&m!=0)cnt++;
    }
    int tmp,i = 1;
    while (cnt) {
        if (mhash[0] == 0) {
            //由于在本位的数我们不动，则只需要让i递增即可
            while (i < n) {
                if (mhash[i] != i) {
                    mhash[0] = mhash[i];
                    mhash[i] = 0;
                    cnt++;
                    break;
                }
                i++;
            }
        }
        tmp = mhash[0];
        mhash[0] = mhash[mhash[0]];
        mhash[tmp] = tmp;
        cnt--;
        count++;
    }
    printf("%d", count);
    return 0;
}