poj 2299 Ultra-QuickSort(求逆序数:離散化+樹状配列)

Ultra-QuickSort
Time Limit: 7000MS
 
Memory Limit: 65536K
Total Submissions: 22147
 
Accepted: 7897
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

 
逆の序数を探して何个があって、逆の序数はあの1つの序列の中で、右の数は左の数より大きい个数の和です.実は泡を探す回数と同じですが、泡を出すことで、50万個のデータが直接死亡します.この問題は木の形の配列でできます.前にはできません.問題解決報告書に木の形の配列ができると書いてみました.そう考えると木の配列が簡単に解決できます.
しかしもう一つの問題は、彼が入力したデータの値が0≦a[i]≦99999999と大きいことですが、データの数はあまり大きくありませんn<500000なので、離散化して解決します!!!
リンク:http://poj.org/problem?id=2299
コード:

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <algorithm>
using namespace std;

int b[500005], c[500005];
int n;

struct node
{
    int num, id;
}a[500005];

bool cmp(node a, node b)
{
    return a.num < b.num;
}

void update(int i, int x)
{
    while(i <= n)
    {
        c[i] += x;
        i += i&(-i);
    }
}

int sum(int i)
{
    int sum = 0;
    while(i > 0)
    {
        sum += c[i];
        i -= i&(-i);
    }
    return sum;
}

int main()
{
    int i;
    long long ans;
    while(scanf("%d", &n), n)
    {
        memset(b, 0, sizeof(b));
        memset(c, 0, sizeof(c));
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &a[i].num); //    num
            a[i].id = i;            //    id
        }
        ///     
        sort(a+1, a+n+1, cmp);  //   
        /*  a[1].num    ，id     ，  b[a[1].id]=1  ，
                1，      2，           */
        b[a[1].id] = 1;
        for(i = 2; i <= n; i++)
        {
            if(a[i].num != a[i-1].num)
                b[a[i].id] = i;
            else b[a[i].id] = b[a[i-1].id];
        }
        ///     
        ans = 0;
        for(i = 1; i <= n; i++)
        {
            update(b[i], 1);
            //     ，      ，               
            ans += (sum(n)-sum(b[i]));
        }
        //    ：                
        printf("%I64d
", ans);
    }

    return 0;
}