PAT A1067. Sort with Swap(0,*) (25)
PAT A1067. Sort with Swap(0,*) (25)
阿豪boy 发表于10个月前
PAT A1067. Sort with Swap(0,*) (25)
  • 发表于 10个月前
  • 阅读 1
  • 收藏 0
  • 点赞 0
  • 评论 0

腾讯云 新注册用户 域名抢购1元起>>>   

https://www.patest.cn/contests/pat-a-practise/1067

 

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

解题思路

首位为0,找出第一个未排序的数的位置进行交换 
首位非0,与该数应处的位置交换

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> a, b;
int main() {
	int n, t, c = 0;
	scanf("%d", &n);
	for (int i = 0; i < n; ++i) {
		scanf("%d", &t);
		a.push_back(t);
	}
	int cnt = 0;
	while (c < n) { //序列排序未结束
		if (0 == a[0]) { //首位为0,找出第一个未排序的数的位置
			for (; c < n && c == a[c]; ++c)
				continue;
			if (c >= n) break;
			swap(a[0], a[c]);
			++cnt;
		} else { //首位非0,与应处于的位置交换
			swap(a[0], a[a[0]]);
			++cnt;
		}
	}
	printf("%d\n", cnt);
	return 0;
}

 

其他参考

http://blog.csdn.net/apie_czx/article/details/48243647

共有 人打赏支持
粉丝 3
博文 466
码字总数 352998
×
阿豪boy
如果觉得我的文章对您有用,请随意打赏。您的支持将鼓励我继续创作!
* 金额(元)
¥1 ¥5 ¥10 ¥20 其他金额
打赏人
留言
* 支付类型
微信扫码支付
打赏金额:
已支付成功
打赏金额: