K.Bro Sorting
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
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 ≤ 10
6).
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 × 10 6.
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 × 10 6.
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: 1HintIn the second sample, we choose “5” so that after the first round, sequence becomes “1 2 3 4 5”, and the algorithm completes.
题目大意：给一个1~n的一种排列，每次随机选一个数，如果下一个数比他小，则交换，一直进行上述过程直到下一个数比他大。最少经过多少次这样的循环能将其变成升序？
大致思路：
想要这样的循环次数最少必定是 每次选最大的不符合位置的数，将其交换至正确的位置。
从后面扫一遍数组，如果当前的位置的数不符合其位置，则循环次数+1，标记当前的数，假设其已移动到正确的位置
#include <cstdio>
#include <cstring>
using namespace std;
int a[1000005];
bool vis[1000005];
int main() {
int T,kase=0,i,n,ans;
scanf("%d",&T);
while(T--) {
scanf("%d",&n);
for(i=1;i<=n;++i)
scanf("%d",a+i);
memset(vis,false,sizeof(vis));
for(ans=0,--i;n;--i) {
if(!vis[a[i]]) {
if(a[i]!=n) {
vis[n]=true;
++ans;
++i;
}
--n;
}
}
printf("Case #%d: %d\n",++kase,ans);
}
return 0;
}