poj 2299

原創

2020-02-21 16:55

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 48121		Accepted: 17553

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

Sample Output

6
0

Source

Waterloo local 2005.02.05

題意：

給出n個數，要求求出按照冒泡排序法需要多少次才能把這n個數按從小到大的順序排列。

思路：

需要多少次才能按從小到大的順序排列，可以轉化成求其逆序數（這個線代學過），但是n<=500000,逆序數可能會超過int ,所以結果要用long long ，否則wrong answer.求其逆序數可以用歸併排序，按照分治三步:

1.劃分問題：吧序列分成元素個數儘量相等的兩半。

2.遞歸求解：把元素分別排列。

3.把兩個有序表合併成一個。

中間過程需要引入一個輔助空間T[n].

每一次累加cnt+=m-p;即是答案。

代碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
int  A[500004],T[500004];
long long  cnt;
using namespace std;
void init(int n)
{
    for(int i=0;i<n;i++)
        scanf("%d",&A[i]);

}
void merge_sort(int *A,int x,int y,int *T)
{
    if(y-x>1)
    {
        int m=x+(y-x)/2;
        int p=x,q=m,i=x;
        merge_sort(A,x,m,T);
        merge_sort(A,m,y,T);
        while(p<m||q<y)
        {
            if(q>=y||(p<m&&A[p]<=A[q]))    T[i++]=A[p++];
            else
            {
                T[i++]=A[q++];
                cnt+=m-p;

            }

    }            for(i=x;i<y;i++)
                A[i]=T[i];
}
}
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
       init(n);
       cnt=0;
     memset(T,0,sizeof(T));
     merge_sort(A,0,n,T);//不能把n寫成n-1，否則會答案錯誤
   printf("%I64d\n",cnt);
    }

    return 0;
}