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

题目大意

用交换的方式将给定的序列排序，要求输出最少的交换次数。

解题思路

使用贪心法计算交换次数：

1. 读入序列，读入序列过程中记录有多少错位的元素；
2. 开始交换过程，当0位置的元素为0时（恰在其位置），寻找到第一个错位的元素与0位置交换；
3. 当0的位置不为0时，则交换0位置与当前0位置元素所应存放位置；
4. 交换过程中计数，直到错位元素数为0；
5. 输出结果并返回零值。

代码

#include<stdio.h>
#define maxn 100010
int pos[maxn];
void Swap(int a,int b){
    int temp;
    temp=pos[a];
    pos[a]=pos[b];
    pos[b]=temp;
}

int main(){
    int N,left,count=0;
    int i,v,k=0;
    scanf("%d",&N);
    left=N-1;
    for(i=0;i<N;i++){
        scanf("%d",&v);
        pos[v]=i;
        if(pos[i]==i&&i!=0){
            left--;
        }
    }
    while(left>0){
        if(pos[0]==0){
            while(k<N){
                k++;
                if(pos[k]!=k){
                    Swap(0,k);
                    count++;
                    break;
                }
            }
        }else{
            Swap(0,pos[0]);
            count++;
            left--;
        }
    }
    printf("%d\n",count);
    return 0;
}

运行结果