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

最新推荐文章于 2020-07-27 10:19:37 发布

walter1990

最新推荐文章于 2020-07-27 10:19:37 发布

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本文探讨了仅使用特殊的Swap(0,*)操作对由0到N-1的整数排列进行排序的问题，并提供了一种算法来确定完成排序所需的最小交换次数。

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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:

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <set>
using namespace std;

int num[100010], myIndex[100010], flag = 0;

void mySwap(int num[], int a, int b) {
	int tmp = num[a];
	num[a] = num[b];
	num[b] = tmp;
}

void output(int num[], int n) {
	for (int i = 0; i < n; ++i)
	{
		printf("%d ", num[i]);
	}
	printf("\n");
}

int main() {
	int n, cnt = 0, i;
	scanf("%d", &n);

	for(int i = 0; i < n; i++) {
		scanf("%d", &num[i]);
		myIndex[num[i]] = i;
		if(num[i] != i)
			flag += 1;
	}
	
	int first = 1;

	while(true) {
		int c1 = myIndex[0];
		if(c1 != 0) {

			int c2 = myIndex[c1];
			mySwap(num, c1, c2);
			myIndex[0] = c2;
			myIndex[c1] = c1;

		} else {
			for(i = first; i < n; i++) {
				if(num[i] != i) {
					myIndex[0] = i;
					myIndex[num[i]] = 0; 
					mySwap(num, 0, i);
					first = i;
					break;
				}
			}

			if(i >= n) break;
		}
		cnt += 1;
	}

	printf("%d\n", cnt);
	return 0;
}