Ultra-QuickSort（离散化）

题目链接：http://poj.org/problem?id=2299

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 75831		Accepted: 28402

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

 

题意:找到给定数组中有多少逆序对

思路：树状数组离散化，存一个找一次 不需要全部存进去再找（这样是找不出来的） 存的时候就找，这样就知道有多少个是逆序的了

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
const LL INF=1e9+7;
const int maxn=5e5+50;
LL a[maxn],b[maxn],c[maxn];
LL len;
LL lowbit(LL x)
{
    return x&(-x);
}
LL query(LL x)
{
    LL ret=0;
    while(x>0)
    {
        ret+=c[x];
        x-=lowbit(x);
    }
    return ret;
}
void update(LL x)
{
    while(x<=500000)
    {
        c[x]++;
        x+=lowbit(x);
    }
}
int main()
{
    LL N;
    while(scanf("%lld",&N)!=EOF)
    {
        if(N==0) break;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        LL ans=0;
        for(int i=1;i<=N;i++)
        {
            scanf("%lld",&a[i]);
            b[i]=a[i];
        }
        sort(b+1,b+N+1);
        len=unique(b+1,b+N+1)-b;
        for(int i=1;i<=N;i++)
        {
    //        cout<<"*"<<endl;
            LL x=lower_bound(b+1,b+len+1,a[i])-b;//找到a[i]在b数组中所在位置
            ans+=i-query(x-1)-1;//query找在x之前的数有多少个  这些都是顺序对的
            update(x);
        }
        printf("%lld
",ans);
    }

    return 0;
}