【PAT】1067. Sort with Swap(0,*) (25)

最新推荐文章于 2024-08-13 22:28:01 发布

realxuejin

最新推荐文章于 2024-08-13 22:28:01 发布

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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:

题意：每次只能用0与另一个数字交换，求将数组变换成升序数组，需要多少次交换操作。

解题思路：（1）假如数字0所处的位置pos != 0, 那么将数字0与原本应该在pos上面的数字交换；（2）假如数字0所处的位置pos=0,那么进行扫描，找出第一个位置错误的数字，将数字0与其交换，然后又回到步骤（1）。详细见下面代码。PS：代码中如果用vector操作会超时，改为数组后可以顺利通过，以下为正确AC的代码。

#include <iostream>
#include <vector>
 
using namespace std; 

struct node{
	int pos;
	int val;
};

int findWrong(int * positions, int begin, int end){
	int i = begin;
	for(; i<end; i++){
		if(positions[i] != i){		
			return i;
		}
	}	
	return 0;
}

int main(int argc, char** argv) {
	int n, i, j, val;
	scanf("%d",&n);
	//vector<int> positions(n);
	int *positions = new int[n];
	vector<bool> flag(n,false);
	for(i=0; i<n; i++){
		scanf("%d",&val);
		positions[val] = i;
	}	
	int step = 0, pos;
	int index;
	index = findWrong(positions, 1, n);
	while(index != 0) { 
		pos = positions[0];
		if(pos != 0){
			//0未归位
			positions[0] = positions[pos] ;
			positions[pos] = pos;
			step++;
		}else{
			//0归位了
			positions[0] = positions[index];
			positions[index] = 0;
			step++;
		}			 
		index = findWrong(positions,index,n);
	}
	printf("%d\n",step);
	return 0;
}