POJ 2299 Ultra-QuickSort

DescriptionIn 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.InputThe 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.OutputFor 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

这个题的意思是问你冒泡排序中交换值的次数，也就是让你求序列中的逆数对之和；

对于这个问题，首先想到的是归并排序求逆数对，在二路归并的时候通过比较左右两边队头的值求得

如归并时：4 5 6 和 1 2 3 ,sum=op[1]+op[2]+op[3]=9;

这个就不多说了；

这个题也可以用树状数组来做：

首先离散化:把输入进去的数用它们所在的下标表示，排序以后它们下标的逆数就代替了它的逆数，这样的好处是使树状数组的下标之间不会差太远，起到一个压缩作用；

其后就是计算了：从数组的最后一个往第一个算，如果它不是最大的，则它到根结点的逆数都要+1；

下面是我用归并按自己理解写的,虽然写的跟挫，但是好懂：

LANGUAGE:C++

CODE:

#include<iostream>using namespace std;long long  cnt;void merge(int array[],int left,int mid,int right){	int* temp=new int[right-left+1];	int i,j,p;	for(i=left,j=mid+1,p=0;i<=mid&&j<=right;p++)	{		if(array[i]<=array[j])temp[p]=array[i++];		else temp[p]=array[j++],cnt+=(mid-i+1);	}	while(i<=mid)temp[p++]=array[i++];	while(j<=right)temp[p++]=array[j++];	for(i=left,p=0;i<=right;i++)array[i]=temp[p++];	delete temp;}void mergesort(int array[],int left,int right){	if(left==right)array[left]=array[right];	else	{		int mid=(left+right)/2;		mergesort(array,left,mid);		mergesort(array,mid+1,right);		merge(array,left,mid,right);	}}int main(){    int n,array[500005];    while(cin>>n,n)    {        cnt=0;        for(int i=0;i<n;i++)            cin>>array[i];        mergesort(array,0,n-1);            cout<<cnt<<endl;    }    return 0;}

这个是用数状数组写的，也不比归并快多少，也是NLOGN的算法；

LANGUAGE:C++

CODE:

#include<iostream>#include<string.h>#include<algorithm>using namespace std;#define maxn 500001struct node{    int key;    int to;    bool const operator<(const struct node& a)const    {        return key<a.key;    }} data[maxn];void update(int interval[],int n,int x,int w){    for(int i=x; i<=n; i+=(i&-i))    {        interval[i]+=w;    }}int getsum(int interval[],int x){    int sum=0;    for(int i=x; i>0; i-=(i&-i))        sum+=interval[i];    return sum;}int main(){    int n,interval[maxn];    long long sum;    while(cin>>n,n)    {        for(int i=1; i<=n; i++)        {            cin>>data[i].key;            data[i].to=i;        }        sum=0;        memset(interval,0,sizeof(interval));        sort(data+1,data+n+1);        for(int i=n; i>0; i--)        {            sum+=getsum(interval,data[i].to);            update(interval,n,data[i].to,1);        }        cout<<sum<<endl;    }}