UVA10570 Meeting with Aliens数学计算

Posted 2022-04-28 海岛Blog

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 N indicating 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
4
1 2 3 4
4
4 3 2 1
4
2 3 1 4
0
Sample Output
0
0
1

问题链接：UVA10570 Meeting with Aliens
问题简述：（略）
问题分析：数学计算问题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10570 Meeting with Aliens */

#include <bits/stdc++.h>

using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 500 + 1;
int n, a[N + N], vis[N];

int getans(int t[])

    memset(vis, 0, sizeof vis);

    int cnt = 0;
    for (int i = 1; i <= n; i++)
        if (!vis[i]) 
            cnt++;
            for (int j = i; !vis[j]; j = t[j])
                vis[j] = 1;
        
    return n - cnt;


int solve()

    int ans = INF;
    for (int i = 1; i <= n; i++)
        ans = min(ans, getans(a + i));

    int l = 1, r = n;
    while (l < r) swap(a[l++], a[r--]);

    for (int i = 1; i <= n; i++)
        a[n + i] = a[i];
    for (int i = 1; i <= n; i++)
        ans = min(ans, getans(a + i));

    return  ans;


int main()

    while (~scanf("%d", &n) && n) 
        for (int i = 1; i <= n; i++) 
            scanf("%d", &a[i]);
            a[n + i] = a[i];
        

        printf("%d\\n", solve());
    

    return 0;