HDU 5122.K.Bro Sorting

349

题目



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 ≤ 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: 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.


题解


求最少要将几个数往后移动到没有比它小能使数组排好序
最初看到题,觉得是最长上升子序列
使了下都是 TLE

换种思路理解题意,计算比后面数大的数的个数

从后往前扫,如果当前数比最前数小,那么记录最小数,否则就将计数器加一
最后计数器记下的数就是答案

这样可以用 O(n) 的时间算出答案,只需要维护一个变量记录最小的数,一个变量计数即可

代码


/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい?
エリーチカ!
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
using namespace std;

const int maxn = 1000005;

int a[maxn];


inline int read_int() {
    char c;
    bool fs = false;
    int ans = 0;
    while(c = getchar(),!(c >= '0'&&c <= '9'))
        if(c == '-')
            fs = true;
    while(c >= '0'&&c <= '9') {
        ans *= 10;
        ans += (int)c - '0';
        c = getchar();
    }
    return fs ? -ans : ans;
}

void Do() {
    int n = read_int();
    int Max = 0;
    bool flag = false;

    for(int i = 1;i <= n;i++) {
        a[i] = read_int();
    }
    int ans = 0;
    int Min = n + 1;
    for(int i = n;i >= 1;i--)
        if(Min > a[i])
            Min = a[i];
        else
            ans++;
    printf("%d\n",ans);
}

int main() {
    int T;
    scanf("%d",&T);
    for(int i = 1;i <= T;i++) {
        printf("Case #%d: ",i);
        Do();
    }
    return 0;
}
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