树状数组离散化(求逆序数)

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  Ultra-QuickSort

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Time Limit: 2.0 Seconds    Memory Limit: 65536K

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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.

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.

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

Output for Sample Input

6
0

Source:  Waterloo Local Contest Feb. 5, 2005

Problem ID in problemset: 1455

数的范围很大，但是只有500000个，学习了一下离散化

用一个struct存值和顺序，然后离散化的时候使用顺序就不会超过范围了。

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
long long int c[500500];
int n;
struct s{
    int v,order;
}a[500500];
int aa[500500];
int lb(int i)
{
    return i&(-i);
}
int get(int i)
{
    int ans=0;
    while(i>0)
    {
        ans+=c[i];
        i-=lb(i);
    }
    return ans; //有时候忘记写返回值，很蠢
}
void sets(int i,int x)
{
    while(i<=n)
    {
        c[i]+=x;
        i+=lb(i);
    }
}
bool cmp(s a,s b)
{
    if(a.v<b.v)
        return 1;
    return 0;
}
int main()
{
    while(cin>>n)
    {
        memset(c,0,sizeof(c));
        long long int ans=0;
        if(n==0)
            break;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i].v;
            a[i].order=i;
        }
        sort(a+1,a+n+1,cmp);
        for(int i=1;i<=n;i++)
            aa[a[i].order]=i;
        for(int i=1;i<=n;i++)
        {
            sets(aa[i],1);
            ans+=(i-get(aa[i]));
        }
        cout<<ans<<endl;
    }
    return 0;
}