acm hnu Problem 10067 Sorting by Swapping

最新推荐文章于 2022-02-12 21:55:34 发布

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最新推荐文章于 2022-02-12 21:55:34 发布

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分类专栏： Algorithm 文章标签： sorting permutation integer output numbers input

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本文详细探讨了ACM湖南大学（HNU）竞赛中的Problem 10067，该问题涉及通过交换操作对整数序列进行排序。文章深入分析了如何有效地实现这一排序策略，同时讲解了输入输出格式和处理整数数组的技巧。

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Sorting by Swapping

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB

Total submit users: 110, Accepted users: 104

Problem 10067 : No special judgement

Problem description

Given a permutation of numbers from 1 to n, we can always get the sequence 1, 2, 3, ..., n by swapping pairs of numbers. For example, if the initial sequence is 2, 3, 5, 4, 1, we can sort them in the following way:

2 3 5 4 1
1 3 5 4 2
1 3 2 4 5
1 2 3 4 5

Here three swaps have been used. The problem is, given a specific permutation, how many swaps we needs to take at least.

Input

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each case contains two lines. The first line contains the integer n (1 <= n <= 10000), and the second line gives the initial permutation.

Output

For each test case, the output will be only one integer, which is the least number of swaps needed to get the sequence 1, 2, 3, ..., n from the initial permutation.

Sample Input

Sample Output

0
3

Problem Source

Beijing 2005 Warmup

#include < stdio.h >
#define MAX 10000
#define RIGHT 1
#define WRONG 0

main()
{
     int total = 0 ,n = 0 ,num[MAX],min = 0 ;
     int i = 0 ,j = 0 ,k = 0 ,p = 0 ,count = 0 ;
     int flag,start;
    scanf( " %d " , & total);

     for ( k = 0 ; k < total ; k ++ ){
        count = 0 ;
        scanf( " %d " , & n);
         for ( p = 0 ; p < n ; p ++ )scanf( " %d " , & num[p]);
         // for( p = 0 ; p < n ; p++)printf("num%d ",num[p]);

         for ( i = 0 ; i < n ; i ++ ){

             if ( num[i] != i + 1 ){
                start = i;
                flag = WRONG;
                 while ( flag != RIGHT ){
                     for ( j = 0 ; j < n ; j ++ ){
                         if ( num[j] == start + 1 ){
                            num[j] = num[start];
                            num[start] = start + 1 ;
                            count ++ ;
                            start = j;
                             break ;
                        }
                    }

                     if ( num[start] == start + 1 )flag = RIGHT;
                     else flag = WRONG;
                } // while
            }
        }
        printf( " %d " ,count);
    }
return 0 ;
}