POJ2299:Ultra-QuickSort(树状数组求逆序数)

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.

5
9
1
0
5
4
3
1
2
3
0

6
0

Waterloo local 2005.02.05

题意：

通过交换来排序，求交换次数，其实就是求逆序数

思路：

用树状数组是一种方法，对于值比较大，我们要先使用离散化

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <list>
#include <algorithm>
#include <climits>
using namespace std;

#define lson 2*i
#define rson 2*i+1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 500005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
const int mod = 1e9+7;

struct node
{
    int x,id;
}a[N];
int n,c[N],r[N];

int cmp(node a,node b)
{
    if(a.x!=b.x)
    return a.x<b.x;
    return a.id<b.id;
}

int sum(int x)
{
    int ret=0;
    while(x>0)
    {
        ret+=c[x];
        x-=lowbit(x);
    }
    return ret;
}

void add(int x,int d)
{
    while(x<=n)
    {
        c[x]+=d;
        x+=lowbit(x);
    }
}

int main()
{
    int i,j,k;
    while(~scanf("%d",&n),n)
    {
        for(i = 1;i<=n;i++)
        {
            scanf("%d",&a[i].x);
            a[i].id = i;
        }
        sort(a+1,a+1+n,cmp);
        MEM(c,0);
        for(i = 1;i<=n;i++)
        {
            r[a[i].id] = i;
        }
        LL ans = 0;
        for(i = 1;i<=n;i++)
        {
            add(r[i],1);//按照输入的顺序来插入，首先对于目前输入的值，统计是否有比其小的值已经插入了，然后再用i减去这些数据，可以得到逆序数
            ans+=(i-sum(r[i]));
        }
        printf("%I64d\n",ans);
    }

    return 0;
}