PAT (Advanced Level) Practise 1067 Sort with Swap(0,*) (25)

本文探讨了一道有趣的算法题目,要求仅通过交换数组中0与其他元素来对0至N-1的排列进行排序,并给出了一种有效的解决方案。

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

时间限制
150 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

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

题意:给你n个数,分别为0~n-1,每次可以将0和其他数字进行交换,问最少要交换几次,使得n个数字最后的顺序为0~n-1

解题思路:若0不在0号位,则将应该在0所在位置的数移过来,直至0到0号位,然后判断所有数所在的位置是否正确,若有不正确的,则用0和它交换


#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[100009], n,x;

int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 0; i < n; i++) scanf("%d", &x),a[x]=i;
		int ans = 0,k=0;
		while (1)
		{
			while (a[0] != 0)
			{
				ans++;
				x = a[0];
				swap(a[x], a[0]);
			}
			int flag = 1;
			for (int i = k; i < n; i++)
			{
				if (i != a[i])
				{
					flag = 0; ans++;
					swap(a[i], a[0]);
					k = i;break;
				}
			}
			if (flag) { printf("%d\n", ans); break; }
		}
	}
	return 0;
}

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