杭电5122

K.Bro Sorting

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Total Submission(s): 520    Accepted Submission(s): 285

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 .

 

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 a_i (1 ≤ a_i ≤ 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

255 4 3 2 155 1 2 3 4

 

Sample Output

Case #1: 4Case #2: 1
Hint
In the second sample, we choose “5” so that after the first round, sequence becomes “1 2 3 4 5”, and the algorithm completes.
 

 

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5122

题意：求题目完成排序需要题目所述的最小的round次数

分析：每次把不符合排序的最大的数进行swap，那么，这个数在经过一个round之后，所有大于等于它的数一定是最终的排列。

由此，可以将问题转化为判断一个数的右边是否有必该数小的数，若有，则需要一次round。

对于此问题，只需要从右往左统计，并且不断更新最小值，若当前数为最小值，则将最小值更新为当前数，否则round+1

注意：用c++写会因为输入和输出超时。

#include <stdio.h>int a[1000005];int main(){    int T, n,i;    scanf("%d", &T);    for(int ca = 1; ca <= T; ca++)    {        scanf("%d", &n);        for( i = 1; i <= n; i++)            scanf("%d", &a[i]);        int mi = a[n], ans = 0;        for( i = n - 1; i >= 1; i--)        {            if(a[i] > mi)                 ans++;            else                 mi = a[i];        }        printf("Case #%d: %d\n", ca, ans);    }}