poj2299 Ultra-QuickSort（树状数组求逆序数，离散化）

所谓数列的逆序数就是一个值sum，sum=b[0]+b[1]+...+b[n-1]。这里假设数列为a，其中b[i]表示在数列中在a[i]后面并且比a[i]小的数的个数。比如有数列 2 8 0 3的逆序数就是1+2+0+0=3。

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

#include<iostream>#include<stdio.h>#include<algorithm>#include<memory.h>using namespace std;const int Max=500010;struct node{    int index;    int v;};node we[Max];int b[Max];int c[Max];int n;bool cmp(node a,node b){    return a.v<b.v;}int lowbit(int i){    return i&(-i);}void update(int i,int x){    while(i<=n)    {        c[i]+=x;        i+=lowbit(i);    }}int Sum(int i){    int sum=0;    while(i>0)    {        sum+=c[i];        i-=lowbit(i);    }    return sum;}int main(){    while(cin>>n&&n)    {        for(int i=1;i<=n;i++)        {            scanf("%d",&we[i].v);            we[i].index=i;        }        sort(we+1,we+n+1,cmp);        b[we[1].index]=1;        for(int j=2;j<=n;j++)//离散化（数据变小，但相对大小不变）        {            if(we[j].v==we[j-1].v)                b[we[j].index]=b[we[j-1].index];            else                b[we[j].index]=j;        }        memset(c,0,sizeof(c));        long long num=0;        for(int k=1;k<=n;k++)//树状数组求逆序数        {            update(b[k],1);            num+=Sum(n)-Sum(b[k]);        }        cout<<num<<endl;    }    return 0;}