Ultra-QuickSort

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

Source

Waterloo local 2005.02.05     本题就是求逆序数   一种方法是用归并排序求逆序数  

/*
POJ 2299 Ultra-QuickSort
求逆序数
用归并排序求逆序数
*/
#include <stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
const int MAXN=500010;
long long ans;//存放逆序数，比较大，必须用long long
int a[MAXN],b[MAXN],c[MAXN];

//将已经排好序的left~mid,mid+1~right进行归并
void merge(int *a,int left,int mid,int right)
{
    int i,j;
    i=0;
    for(j=left;j<=mid;j++)
       b[i++]=a[j];
    int len1=mid-left+1;
    i=0;
    for(j=mid+1;j<=right;j++)
       c[i++]=a[j];
    int len2=right-mid;
    i=0;
    j=0;
    int k=left;
    while(i<len1&&j<len2&&k<=right)
    {
        if(b[i]<=c[j])
        {
            a[k++]=b[i++];
        }
        else
        {
            a[k++]=c[j++];
            ans+=(len1-i);//逆序数就是累加后面比自己小的数的个数
            //此时b[i]>c[j],那么c[j]会给b[i]后面的len1-i个数造成逆序数
        }
    }
    while(i<len1) a[k++]=b[i++];
    while(j<len2) a[k++]=c[j++];
}
void merge_sort(int *a,int left,int right)//对a[left~right-1]进行归并排序
{
    if(left<right)
    {
        int mid=(left+right)/2;
        merge_sort(a,left,mid);
        merge_sort(a,mid+1,right);
        merge(a,left,mid,right);
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    while(scanf("%d",&n),n)
    {
        for(int i=0;i<n;i++)
          scanf("%d",&a[i]);
        ans=0;
        merge_sort(a,0,n-1);
        printf("%I64d\n",ans);
    }
    return 0;
}

另外一种求逆序数的方法就是用树状数组。

本题数据比较大，需要先离散化，再用树状数组。

代码如下：

/*
POJ 2299 Ultra-QuickSort
求逆序数
离散化+树状数组
首先从小到大进行编号，从而实现离散化
然后利用树状数组来统计每个数前面比自己大的数的个数
*/

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=500010;
int c[MAXN];
int b[MAXN];
int n;
struct Node
{
    int index;//序号
    int v;
}node[MAXN];
bool cmp(Node a,Node b)
{
    return a.v<b.v;
}
int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int val)
{
    while(i<=n)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}
int main()
{
   // freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%d",&n),n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&node[i].v);
            node[i].index=i;
        }
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        //离散化
        sort(node+1,node+n+1,cmp);
        //将最小的编号为1
        b[node[1].index]=1;
        for(int i=2;i<=n;i++)
        {
            if(node[i].v!=node[i-1].v) b[node[i].index]=i;
            else b[node[i].index]=b[node[i-1].index];
        }
        long long  ans=0;
        //这里用的很好
        //一开始c数组都是0，然后逐渐在b[i]处加上1；
        for(int i=1;i<=n;i++)
        {
            add(b[i],1);
            ans+=sum(n)-sum(b[i]);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}