HDU 5122 K.Bro Sorting 2014ICPC 北京站现场赛

K.Bro Sorting

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1249    Accepted Submission(s): 563

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.
 

 

题意：给你1~n的n个数，每次轮随机选择一个数，然后将这个数移动到比他大的数的前一个位置，求最小的轮数。

分析：每次选择最大的移动即可，从后往前找，如果当前数后面有比它小的，那么它肯定需要移动，就这样每次维护当前数后面最小的数，如果当前数比它大轮数就加1，否则更新最小值。

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <cctype>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-8)#define inf 0x3f3f3f3f#define ll long long int#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1using namespace std;const int mod = 1000000007;const int Max = 1000005;int t,n;int a[Max];int main(){    int cnt=1;    cin>>t;    while(t--)    {        scanf("%d",&n);        int ans=0;        a[0]=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        int m=inf;        for(int i=n;i>=1;i--)        {            if(a[i]>m)                ans++;            m=min(m,a[i]);        }        printf("Case #%d: %d\n",cnt++,ans);    }    return 0;}

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