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  • POJ 2299 Ultra-QuickSort

    Ultra-QuickSort
    Time Limit: 7000MS   Memory Limit: 65536K
    Total Submissions: 39767   Accepted: 14336

    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

    5
    9
    1
    0
    5
    4
    3
    1
    2
    3
    0
    

    Sample Output

    6
    0
    



    归并排序:

    #include<iostream>
    #include<cstring>
    using namespace std;
    
    #define M 500005
    int a[M],b[M];
    __int64 ans;
    
    void mergr_sort(int x,int y)
    {
    	if(y-x>1)
    	{
    		int m=x+(y-x)/2;   //划分
    		int p=x,q=m,i=x;
    		mergr_sort(x,m);    //递归求解
    		mergr_sort(m,y);    //递归求解
    		while(p<m || q<y)
    		{
    			if(q>=y || (p<m && a[p]<=a[q]))
    				b[i++]=a[p++];     //把左半数组拷贝到暂时空间
    			else
    			{
    				b[i++]=a[q++];     //把右半数组拷贝到暂时空间
    				ans+=m-p;    
    			}
    		}
    		for(i=x;i<y;i++) a[i]=b[i];
    	}
    }
    
    int main()
    {
    	int n;
    	while(cin>>n)
    	{
    		if(n==0)  break;
    		memset(a,0,sizeof(a));
    		memset(b,0,sizeof(b));
    		int i;
    		for(i=1;i<=n;i++)
    			cin>>a[i];
    
    		ans=0;
    		mergr_sort(1,n+1);
    		printf("%I64d
    ",ans);
    	}
    
    	return 0;
    }
    
    
    
    
    
    
    
    


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  • 原文地址:https://www.cnblogs.com/gccbuaa/p/6883172.html
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