Ultra-QuickSort(裸树状数组求逆序数)

题目描述

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.

输入

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.

输出

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.

样例输入

5
9
1
0
5
4
3
1
2
3
0

样例输出

6
0

#include <bits/stdc++.h>
#define ll long long
#define met(a) memset(a,0,sizeof(a));
using namespace std;
const int mod=1e9+7;
int s[500010],aa[500010],n;
struct node{int b,c;};
node a[500010];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y)
{
    while(x<=n)
        aa[x]+=y,
        x+=lowbit(x);
}
int sum(int x)
{
    int t=0;
    while(x>=1)
        t+=aa[x],x-=lowbit(x);
    return t;
}
bool cmp(node x,node y)
{
    return x.b<y.b;
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int i,j,k,m;
    ll cnt=0;
    met(a);
    while(cin>>n&&n)
    {
        cnt=0;
        for(i=1;i<=n;i++)
        {
            cin>>a[i].b;
            a[i].c=i;
        }
        sort(a+1,a+1+n,cmp);
        for(i=1;i<=n;i++)
            s[a[i].c]=i;
        met(aa);
        for(i=1;i<=n;i++)
        {
            update(s[i],1);
            cnt+=(i-sum(s[i]));
        }
        cout<<cnt<<endl;
    }
    return 0;
}

View Code