poj 2299 Ultra-QuickSort

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

59105431230

Sample Output

60

题目大意：

求逆序对，方法多种多样，树状数组、线段树、归并都可以做，现在练的是树状数组，所以介绍一下树状数组的解释。

      总共有N个数，如何判断第i+1个数到最后一个数之间有多少个数小于第i个数呢？不妨假设有一个区间 [1,N]，只需要判断区间[i+1,N]之间有多少个数小于第i个数。如果我们把总区间初始化为0，然后把第i个数之前出现过的数都在相应的区间把它的值定为1，那么问题就转换成了[i+1,N]值的总和。再仔细想一下，区间[1,i]的值+区间[i+1,N]的值=区间[1,N]的值(i已经标记为1)，所以区间[i+1,N]值的总和等于N-[1,i]的值！因为总共有N个数，不是比它小就是比它(大或等于)。   
现在问题已经转化成了区间问题，枚举每个数，然后查询这个数前面的区间值的总和，i-[1,i]既为逆序数

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define ll long long#define N 500001struct node{    int val,id;};node p[N];int n,c[N],ha[N];int cmp(node a,node b){    return a.val<b.val;}int lowbit(int x){    return x&(-x);}void update(int x,int val){    while (x<=n)    {        c[x]+=val;        x+=lowbit(x);    }}int sum(int x){    int ans=0;    while (x>=1)    {        ans+=c[x];        x-=lowbit(x);    }    return ans;}int main(){    int i;    while (scanf("%d",&n),n)    {      for (i=1;i<=n;i++)        {            scanf("%d",&p[i].val);            p[i].id=i;        }    sort(p+1,p+n+1,cmp);  //  for (i=1;i<=n;i++)  //      printf("%d %d\n",p[i].id,p[i].val);    for (i=1;i<=n;i++)         ha[p[i].id]=i;   //哈希数组里面代表的意义是第i个数字按顺序最终应放在ha[i]的位置上    memset(c,0,sizeof(c));    ll ans=0;    for (i=1;i<=n;i++)    {        update(ha[i],1);        ans+=(i-sum(ha[i]));    //    }    printf("%lld\n",ans);    }    return 0;}