poj 2299 Ultra-QuickSort（求逆序数：离散化+树状数组）

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

 

        此题大意：找逆序数有多少个，逆序数就是那一个序列中，右边的数比左边的数大的个数之和。其实就是和找冒泡的次数一样，不过用冒泡来做，50万个数据直接死亡，这道题可以用树状数组来做，之前不会做，看到解题报告中有说树状数组可以，就开始尝试写，后来发现竟然可以这样想：每次找前面（左边）的数比自己大的个数的总和！！！这样子想的话树状数组可以轻松解决。

      不过还有一个问题，就是他输入的数据的值很大0 ≤ a[i] ≤ 999,999,999，但是数据的数量不是很大n < 500,000 ，所以用到离散化解决！！！

链接：http://poj.org/problem?id=2299

代码：

#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); //输入数值num
            a[i].id = i;            //记录序号id
        }
        ///开始离散化
        sort(a+1, a+n+1, cmp);  //先排序
        /*因为a[1].num是最小的，id是它的位置，所以b[a[1].id]=1最小，
          最小的数变成1，第二小的变成2，如此类推从而达到离散化*/
        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];
        }
        ///离散化完毕
        ans = 0;
        for(i = 1; i <= n; i++)
        {
            update(b[i], 1);
            //这里很巧妙，每一次更新后，判断此数比左边的数小的数有多少
            ans += (sum(n)-sum(b[i]));
        }
        //从而求到：右边的数比左边的数大的个数的总和
        printf("%I64d\n", ans);
    }

    return 0;
}