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

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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

这里当0处于0号位置时，查找要花费比较多得时间，所以用一个变量记录上次查找到得位置，那么下次查找就从这个记录的位置开始查找就可以了

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#include <cstdio>
#include <algorithm>

using namespace std;

int main() {
    int n = 0;
    scanf("%d", &n);
    int *pos = new int[n];
    int unsort = 0;
    for (int i = 0; i < n; i++) {
        int num = 0;
        scanf("%d", &num);
        pos[num] = i;
        if (i != pos[i] && i != 0) unsort++;
    }
    int cnt = 0, k = 0;
    while (unsort > 0) {
        if (pos[0] != 0) {
            swap(pos[0], pos[pos[0]]);
            cnt++;
            unsort--;
        } else {
            while (k < n) {
                if (k != pos[k]) {
                    swap(pos[0], pos[k]);
                    cnt++;
                    break;
                }
                k++;
            }
        }
    }
    printf("%d\n", cnt);
    delete[] pos;
    return 0;
}