首页 > 代码库 > K.Bro Sorting
K.Bro Sorting
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
Matt’s friend K.Bro is an ACMer.
Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong order. The process repeats until no swap is needed.
Today, K.Bro comes up with a new algorithm and names it K.Bro Sorting.
There are many rounds in K.Bro Sorting. For each round, K.Bro chooses a number, and keeps swapping it with its next number while the next number is less than it. For example, if the sequence is “1 4 3 2 5”, and K.Bro chooses “4”, he will get “1 3 2 4 5” after this round. K.Bro Sorting is similar to Bubble sort, but it’s a randomized algorithm because K.Bro will choose a random number at the beginning of each round. K.Bro wants to know that, for a given sequence, how many rounds are needed to sort this sequence in the best situation. In other words, you should answer the minimal number of rounds needed to sort the sequence into ascending order. To simplify the problem, K.Bro promises that the sequence is a permutation of 1, 2, . . . , N .
Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong order. The process repeats until no swap is needed.
Today, K.Bro comes up with a new algorithm and names it K.Bro Sorting.
There are many rounds in K.Bro Sorting. For each round, K.Bro chooses a number, and keeps swapping it with its next number while the next number is less than it. For example, if the sequence is “1 4 3 2 5”, and K.Bro chooses “4”, he will get “1 3 2 4 5” after this round. K.Bro Sorting is similar to Bubble sort, but it’s a randomized algorithm because K.Bro will choose a random number at the beginning of each round. K.Bro wants to know that, for a given sequence, how many rounds are needed to sort this sequence in the best situation. In other words, you should answer the minimal number of rounds needed to sort the sequence into ascending order. To simplify the problem, K.Bro promises that the sequence is a permutation of 1, 2, . . . , N .
Input
The first line contains only one integer T (T ≤ 200), which indicates the number of test cases. For each test case, the first line contains an integer N (1 ≤ N ≤ 106).
The second line contains N integers ai (1 ≤ ai ≤ N ), denoting the sequence K.Bro gives you.
The sum of N in all test cases would not exceed 3 × 106.
The second line contains N integers ai (1 ≤ ai ≤ N ), denoting the sequence K.Bro gives you.
The sum of N in all test cases would not exceed 3 × 106.
Output
For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), y is the minimal number of rounds needed to sort the sequence.
Sample Input
2
5
5 4 3 2 1
5
5 1 2 3 4
Sample Output
Case #1: 4
Case #2: 1
Hint
In the second sample, we choose “5” so that after the ?rst round, sequence becomes “1 2 3 4 5”, and the algorithm completes.1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 using namespace std; 6 const int ms=1000005; 7 int a[ms]; 8 int n,p,cnt; 9 void input()10 {11 scanf("%d",&n);12 for(int i=0;i<n;i++)13 scanf("%d",&a[i]);14 }15 16 void solve()17 {18 cnt=0;19 int flag=a[n-1];20 for(int i=n-2;i>=0;i--)21 {22 if(a[i]>flag)23 cnt++;24 else25 flag=a[i];26 }27 printf("Case #%d: %d\n",p++,cnt);28 }29 int main()30 {31 int T;32 scanf("%d",&T);33 p=1;34 while(T--)35 {36 input();37 solve();38 }39 return 0;40 }
K.Bro Sorting
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。