PAT甲级练习1067. Sort with Swap(0,*) (25)

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

只能用零去换其它位置上的数，因此分为零在第零位上和不在第零位上两种情况。不在零位上时，将零与当前位置对应的数进行互换；在零位上时，找到最近的数与位置不相符的位置进行替换，然后重复不在零位上的情况。用一个cnt来统计需要调整的位数。

#include <iostream>  
#include <cstdio>  
#include <algorithm>  
#include <vector>  
#include <map>  
#include <set>  
#include <stack>  
#include <queue>  
#include <string>  
#include <string.h> 
using namespace std;

const int MAX = 1e5+10;
int n, num, an, cnt, ind=1, l[MAX];

int main() {

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

	while(an>0){
		if(l[0]==0){//零在第零个位置
			while(ind<n){
				if(l[ind]!=ind){
					swap(l[ind], l[0]);
					cnt++;
					break;
				}
				ind++;
			}
		}
		while(l[0]!=0){
			swap(l[l[0]], l[0]);
			cnt++;
			an--;
		}
	}
	printf("%d", cnt);

	scanf("%d",&n);
    return 0;
}