Bro Sorting（2014ACM/ICPC亚洲区北京站-K）
来源：程序员人生 发布时间：2014-12-10 08:42:59 阅读次数：3113次
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 .
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.
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.
过的莫名其妙的1道题。看着数据10^6太大，常规方法肯定TLE，然后在想如果是交换的话肯定是后面的数值比前面的数值大，所以就设定第1个为最大的，后面要是出现比他大的就交换，否则的话就就cnt++，然后敲出来第2组数据的结果是4，后来才发现是找到比当前的数大的数就停止。然后就在想倒着来行不行，由于倒着的话没有前面的那个碰到比自己大的数就停止的限制。就令最后的值为最小值，碰到前面比自己大的就cnt++，否则就更新当前最小值。（如果你正着来看的话正好是是处于后面的值比前面的值小，而不是交换）。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int a[1000010];
int main()
{
int T,n,i,j;
int min;
int cnt;
scanf("%d",&T);
for(j=1;j<=T;j++)
{
cnt=0;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
min=a[n⑴];
for(i=n⑵;i>=0;i--)
{
if(min>a[i])
min=a[i];
else
cnt++;
}
printf("Case #%d: %d
",j,cnt);
}
return 0;
}
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