求逆序对——Ultra-QuickSort 题解解析

Description

Inthisproblem,youhavetoanalyzeaparticularsortingalgorithm.Thealgorithmprocessesasequenceofndistinctintegersbyswappingtwoadjacentsequenceelementsuntilthesequenceissortedinascendingorder.Fortheinputsequence

91054,

Ultra-QuickSortproducestheoutput

01459.

YourtaskistodeterminehowmanyswapoperationsUltra-QuickSortneedstoperforminordertosortagiveninputsequence.



Input

Theinputcontainsseveraltestcases.Everytestcasebeginswithalinethatcontainsasingleintegern<500,000--thelengthoftheinputsequence.Eachofthethefollowingnlinescontainsasingleinteger0≤a[i]≤999,999,999,
thei-thinputsequenceelement.Inputisterminatedbyasequenceoflengthn=0.Thissequencemustnotbeprocessed.

Output

Foreveryinputsequence,yourprogramprintsasinglelinecontaininganintegernumberop,theminimumnumberofswapoperationsnecessarytosortthegiveninputsequence

SampleInput

5

91054

3123
0

SampleOutput

6
0

求逆序对的问题可以用冒泡排序，但是当序列够长的时候，冒泡排序的复杂度是不能满足要求的。我们可以用归并排序的思路求解。

代码如下：

#include<iostream>
#include<sstream>
#include<ios>
#include<iomanip>
#include<functional>
#include<algorithm>
#include<vector>
#include<string>
#include<list>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<climits>
#include<cctype>
usingnamespacestd;
#defineXINFINT_MAX
#defineINF0x3F3F3F3F
#defineMP(X,Y)make_pair(X,Y)
#definePB(X)push_back(X)
#defineREP(X,N)for(intX=0;X<N;X++)
#defineREP2(X,L,R)for(intX=L;X<=R;X++)
#defineDEP(X,R,L)for(intX=R;X>=L;X--)
#defineCLR(A,X)memset(A,X,sizeof(A))
#defineITiterator
typedeflonglongll;
typedefpair<int,int>PII;
typedefvector<PII>VII;
typedefvector<int>VI;
//constintMAXN=10010;
//#defineINF0x3FFFFFFF
/*************************************
**************头文件*****************
*************************************/
lla[500005],b[500005];
llans;
voidMerge(lla[],lllow,llmid,llhigh)
{
//	ll*b=newll[high+1];/////不能动态分配内存，会TME
	lli=low,j=mid+1,k=low;
	while(i<=mid&&j<=high)
		if(a[i]<=a[j])
			b[k++]=a[i++];	
		else
		{
			b[k++]=a[j++];
			ans+=mid+1-i;		//这里将a[j]赋给b[k]中间跳过了mid+1-i个数
		}						//这些数就是a[j]共有多少个逆序对
	while(i<=mid)
		b[k++]=a[i++];
	while(j<=high)
		b[k++]=a[j++];
	for(i=low;i<=high;i++)
		a[i]=b[i]；
}
voidMergeSort(lla[],lln)//两路归并
{
	lllen=1,i;
	while(len<n){
		i=0;
		while(i+2*len<=n){
			Merge(a,i,i+len-1,i+2*len-1);
			i+=2*len;
		}
		if(i+len<n)
			Merge(a,i,i+len-1,n-1);
		len*=2;
	}
}
intmain()
{
	lln;
	while(cin>>n&&n)
	{
		REP(i,n)
			cin>>a[i];
		ans=0;
		MergeSort(a,n);					//调用归并函数进行排序
		cout<<ans<<endl;				//直接输出答案
	}
	return0;
}