Ultra-QuickSort (POJ 2299)树状数组+离散化

题目链接

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

题目a[i]范围9e9+9，emmm，似乎会爆空间，似乎写过类似题，似乎必须离散化，似乎就这么写了。。。。。。

然后用树状数组解题并解出逆序数就...巴拉巴拉

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//#include<bits/stdc++.h>
#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <ctype.h>
#include <map>
#include <vector>
#include <set>
using namespace std;
const int maxn=5e5+10;
int Tree[maxn],b[maxn];
struct node
{
    int val,pos;
}a[maxn];
bool cmp(node x,node y)
{
    return x.val<y.val;
}
inline int lowbit(int x) //计算2^k
{
    return (x&-x);
}
void add(int x,int value)
{
    for(int i=x; i<maxn; i+=lowbit(i))
            Tree[i]+=value;
}
int get(int x)
{
    int sum=0;
    for(int i=x; i>0; i-=lowbit(i))
            sum+=Tree[i];
    return sum;
}
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        memset(Tree,0,sizeof(Tree));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i].val);;
            a[i].pos=i;
        }
        sort(a+1,a+n+1,cmp);
        for(int i=1;i<=n;i++) b[a[i].pos]=i;
        long long ans=0;
        for(int i=1;i<=n;i++)
        {
            add(b[i],1);
            ans+=i-get(b[i]);
        }
        printf("%I64d
",ans);
    }
    return 0;
}