Bro Sorting（2014ACM/ICPC亞洲區北京站-K）
來源：程序員人生 發布時間：2014-12-10 08:42:59 閱讀次數：3249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.
過的莫名其妙的1道題。看著數據10^6太大，常規方法肯定TLE，然后在想如果是交換的話肯定是后面的數值比前面的數值大，所以就設定第1個為最大的，后面要是出現比他大的就交換，否則的話就就cnt++，然后敲出來第2組數據的結果是4，后來才發現是找到比當前的數大的數就停止。然后就在想倒著來行不行，由于倒著的話沒有前面的那個碰到比自己大的數就停止的限制。就令最后的值為最小值，碰到前面比自己大的就cnt++，否則就更新當前最小值。（如果你正著來看的話正好是是處于后面的值比前面的值小，而不是交換）。
#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;
}
生活不易，碼農辛苦
如果您覺得本網站對您的學習有所幫助,可以手機掃描二維碼進行捐贈
