10-排序6 Sort with Swap(0, i) （25 分）

标签：one   input   algorithm   ==   imu   sam   NPU   clu   rate   

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

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100010;
int pos[maxn];
int main(){
    int n,ans = 0;
    scanf("%d",&n);
    int left = n - 1,num;
    for(int i = 0; i < n; i++){
        scanf("%d",&num);
        pos[num] = i;
        if(num == i && num != 0){
            left--;
        }
    }
    int k = 1;
    while(left > 0){
        if(pos[0] == 0){
            while(k < n){
                if(pos[k] != k){
                    swap(pos[0],pos[k]);
                    ans++;
                    break;
                }
                k++;
            }
        } 
        if(pos[0] != 0){
            swap(pos[0],pos[pos[0]]);
            ans++;
            left--; 
        }
    }
    printf("%d",ans);
    return 0;
}

10-排序6 Sort with Swap(0, i) （25 分）

标签：one   input   algorithm   ==   imu   sam   NPU   clu   rate   

原文地址：https://www.cnblogs.com/wanghao-boke/p/10409287.html