10570 - Meeting with Aliens（枚举贪心数论）

最新推荐文章于 2020-12-28 11:04:19 发布

lab104_yifan

最新推荐文章于 2020-12-28 11:04:19 发布

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本文链接：https://blog.youkuaiyun.com/accelerator_/article/details/16847859

本文探讨了一个关于序列排列的问题，即在给定一个序列的情况下，通过最少的交换操作来使其变为有序序列的方法。通过暴力枚举每个位置作为起点的正反序列，并利用贪心策略进行操作，最终找出最小的交换次数。该问题适用于理解排序算法和贪心策略的应用。

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Problem D
Meeting with Aliens
Input: Standard Input

Output: Standard Output

Time Limit: 3 Seconds

The aliens are in an important meeting just before landing on the earth. All the aliens sit around a round table during the meeting. Aliens are numbered sequentially from 1 to N. It's a precondition of the meeting that i'th alien will sit between (i-1)'th and (i+1)'th alien. 1st alien will sit between 2nd and N'th alien.

Though the ordering of aliens are fixed but their positions are not fixed. In the above figure two valid sitting arrangements of eight aliens are shown. Right before the start of the meeting the aliens sometimes face a common problem of not maintaining the proper order. This occurs as no alien has a fixed position. Two maintain the proper order, two aliens can exchange their positions. The aliens want to know the minimum number of exchange operations necessary to fix the order.

Input

Input will start with a positive integer, N (3<=N<=500) the number of aliens. In next few lines there will be N distinct integers from 1 to Nindicating the current ordering of aliens. Input is terminated by a case where N=0. This case should not be processed. There will be not more than 100 datasets.

Output

For each set of input print the minimum exchange operations required to fix the ordering of aliens.

Sample Input Output for Sample Input

1 2 3 4

4 3 2 1

2 3 1 4

题意：给定一个序列，求最少几步得到有序序列。

思路：暴力枚举每个位置作为起点的正反序列，然后利用贪心，第i个位置放i，一个个位置去对应，最后枚举出最小的答案

代码：

#include <stdio.h>
#include <string.h>
const int INF = 0x3f3f3f;
const int N = 505;
int min(int a, int b) {return a < b ? a : b;}

int n, num[N], i, j, save[N], v[N];

void getstart1(int j) {
	for (int i = 1; i <= n; i ++) {
		if (j + i - 1 <= n) {
			save[i] = num[j + i - 1];
			v[num[j + i - 1]] = i;
		}
		else {
			save[i] = num[j + i - 1 - n];
			v[num[j + i - 1 - n]] = i;
		}
	}
}

void getstart2(int j) {
	for (int i = 1; i <= n; i ++) {
		if (j - i + 1 > 0) {
			save[i] = num[j - i + 1];
			v[num[j - i + 1]] = i;
		}
		else {
			save[i] = num[j - i + 1 + n];
			v[num[j - i + 1 + n]] = i;
		}
	}
}

int cal() {
	int ans = 0;
	for (int i = 1; i <= n; i ++) {
		if (save[i] != i) {
			save[v[i]] = save[i];
			v[save[i]] = v[i];
			ans ++;
		}
	}
	return ans;
}

void out() {
	for (int i = 1; i < n; i ++)
		printf("%d ", save[i]);
	printf("%d\n", save[i]);
}

int solve() {
	int ans = INF;
	for (int i = 1; i <= n ; i ++) {
		getstart1(i);
		ans = min(ans, cal());
		//out();
		getstart2(i);
		ans = min(ans, cal());
		//out();
	}
	return ans;
}

int main() {
	while (~scanf("%d", &n) && n) {
		for (i = 1; i <= n; i ++)
			scanf("%d", &num[i]);
		printf("%d\n", solve());
	}
	return 0;
}