POJ 2299 Ultra-QuickSort(树状数组+离散化 或 归并排序求逆序)

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

给你一个n个整数组成的序列，每次只能交换相邻的两个元素，问你最少要进行多少次交换才能使得整个整数序列上升有序。

 假设当前处理第i个数,我们只需要计算出i的逆序加到总和ans上即可.i的逆序为:在i之前的那些比i大的数的个数.所以从0到n-1一一扫描,令x[v]=1,表示之前的扫描已经有一个值为v的数被扫描到了.所以当我们处理第i个数a[i]的时候,它的逆序为:x[max]+x[max-1]+…+x[a[i]+1]的值( 即为sum(max)-(x[0]+x[1]+…+x[a[i]-1]) ),且我们需要令x[a[i]]++.

        最终可以算出逆序总值ans.

        但是此题的max高达10亿-1,我们不可能去开一个这么大的数组,但是数只有50W个,我们可以开个50W的数组.,而且我们需要的逆序数仅仅相关与数之间的相对大小,比如3,888,1000000 这三个数的序列我们完全可以用1,2,3这三个数的序列代替,他们的逆序数是一样的.

        所以我们将先对读入的数组离散化处理,使得他们的值集中,但是不影响他们之间的相对大小.

然后再用树状数组即可.

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<queue>
#include<map>
#include<iomanip>
#include<math.h>
using namespace std;
typedef long long ll;
typedef double ld;
const int MAXN=500000+1000;
int c[MAXN];

int lowbit(int x)
{
    return x&(-x);
}

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

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

struct node
{
    int v;
    int index;
    bool operator<(const node& b)const
    {
        return v<b.v;
    }
} nodes[MAXN];

int b[MAXN];//将初始数组重新赋值后 相对大小不变的新数组
int main()
{
    int n;
    while(scanf("%d",&n)==1&&n)
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&nodes[i].v);
            nodes[i].index=i;
        }
        sort(nodes+1,nodes+n+1);
        memset(b,0,sizeof(b));
        b[nodes[1].index]=1;
        for(int i=2; i<=n; i++)
        {
            if(nodes[i].v==nodes[i-1].v)
                b[ nodes[i].index ]=b[ nodes[i-1].index ];
            else
                b[ nodes[i].index ]=i;
        }
        memset(c,0,sizeof(c));
        long long ans=0;
        for(int i=1; i<=n; i++)
        {
            add(b[i],1);//当前扫描的值是b[i],那么在x[b[i]]这个点上加1,表示又出现了1个b[i]值
            ans += sum(n)-sum(b[i]);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}