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

最新推荐文章于 2020-04-21 18:48:59 发布

多一些不为什么的坚持

最新推荐文章于 2020-04-21 18:48:59 发布

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本文探讨了一种特殊的排序问题——仅允许使用Swap(0, *)操作来对0到N-1的排列进行排序。通过巧妙的方法记录每个数的位置，并通过特定的交换策略实现高效排序，最终求出最小交换次数。

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:

思路是在网上看的，主要就是a[num]=i;表示num这个数字在位置i上。然后交换的是位置

这样就避免了寻找0的位置和0归位后第一个不归位的数字

#include <cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
int main() {
    int n;
	cin>>n;
	int a[100001];
	int cnt=0;
	for(int i=0;i<n;i++){
		int num;
		scanf("%d",&num);
		a[num]=i;
		if(a[num]==num) cnt++;
	} 
	
	int sum=0;
	int index=0;
	while(cnt<n-1){
		if(a[0]==0){
			for(;index<n;index++){
				if(index!=a[index]){
					swap(a[0],a[index]);
					sum++;
					break;
				}
			}
		}
		while(a[0]!=0){
			swap(a[0],a[a[0]]);
			sum++;
			cnt++;
		}
	   // cout<<cnt<<endl;
	}
	cout<<sum;
    return 0;
}