1067 Sort with Swap(0, i) (25 分)

1067 Sort with Swap(0, i) (25 分)

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 (≤10
​5
​​ ) 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
通过模拟容易得出以下算法。
尽可能每次swap都让一个数字到它最终的位置，所以
一、当num[0] != 0时,则交换0和等于它位置下标的数。
二、当num[0] == 0时，此时无法通过一个swap让一个数到达它最终的位置，故将0和一个未到达最终位置的数交换位置。
本方法可以得到正确的结果，但提交时出现两个超时。参考网上的资料，感觉像是打开了一个新世界的大门。
使用一个位置数组pos[i] = k,表示数字i的位置为k，好处在于可以快速定位数字的位置，且可以在线性时间复杂度下找到未到达最终位置的数。代码如下：

#include 
#include 
using namespace std;
const int maxn = 100010;
int pos[maxn];  //存储数字在数组中的位置
int main(){
    int n,index = 1,cnt = 0;
    scanf("%d",&n);
    for (int i = 0; i < n; ++i) {
        int temp;
        scanf("%d",&temp);
        pos[temp] = i;
    }
    while(1){
        while(pos[0] != 0){
            swap(pos[0],pos[pos[0]]);	//交换0和pos[0]的位置
            cnt++;
        }
        while (index < n && pos[index] == index)index++;//找到未到最终位置的最小数字
        if(index == n)break;	//循环退出条件
        swap(pos[0],pos[index]);	//交换两个数的位置等同理解为交换两个数
        cnt++;
    }
    printf("%d",cnt);
    return 0;
}