UVA10570-Meeting with Aliens（枚举）

Problem UVA1616-Caravan Robbers

Accept: 531 Submit: 2490
Time Limit: 3000 mSec

Problem Description

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

题解：这个题很有价值。想到倍长数列是比较自然的，但是接下来怎么办，如何快速求出将一个序列排成有序的最小交换次数，这里要用到一个结论：对于一个长度为n的元素互异的序列，通过交换实现有序的最小的交换次数是=n - n被分解成单循环的个数。具体证明见如下博客：

https://blog.csdn.net/wangxugangzy05/article/details/42454111

明白了这个，题目就变得很简单了，枚举起点,dfs找环，取最大值得出结果，这里要注意一点就是序列既可以是升序，也可以是降序，因此要倒着再枚举一遍，方法不变。

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn =  + ;

 int n;

 int num[maxn << ];

 bool vis[maxn];

 void dfs(int st, int a) {

     if (vis[a]) return;

     vis[a] = true;

     dfs(st, num[st + a - ]);

 }

 void dfs2(int st, int a) {

     if (vis[a]) return;

     vis[a] = true;

     dfs2(st, num[st - a + ]);

 }

 int main()

 {

     //freopen("input.txt", "r", stdin);

     while (~scanf("%d", &n) && n) {

         for (int i = ; i < n; i++) {

             scanf("%d", &num[i]);

             num[i + n] = num[i];

         }

         int Max = ;

         for (int st = ; st < n; st++) {

             memset(vis, false, sizeof(vis));

             int cnt = ;

             for (int i = st; i < st + n; i++) {

                 if (!vis[num[i]]) {

                     dfs(st, num[i]);

                     cnt++;

                 }

             }

             Max = Max > cnt ? Max : cnt;

         }

         for (int st =  * n - ; st >= n; st--) {

             memset(vis, false, sizeof(vis));

             int cnt = ;

             for (int i = st; i >= st - n + ; i--) {

                 if (!vis[num[i]]) {

                     dfs2(st, num[i]);

                     cnt++;

                 }

             }

             Max = Max > cnt ? Max : cnt;

         }

         printf("%d\n", n - Max);

     }

     return ;

 }

巴特西

UVA10570-Meeting with Aliens（枚举）

Accept: 531 Submit: 2490
Time Limit: 3000 mSec

Problem Description

Input

Output

Sample Input

Sample Output

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热门文章

巴特西

UVA10570-Meeting with Aliens（枚举）

Accept: 531 Submit: 2490Time Limit: 3000 mSec

Problem Description

Input

Output

Sample Input

Sample Output

最新文章

热门文章

Accept: 531 Submit: 2490
Time Limit: 3000 mSec