树状数组--Ultra-QuickSort

最新推荐文章于 2021-05-01 17:03:48 发布

原创最新推荐文章于 2021-05-01 17:03:48 发布 · 320 阅读

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原题链接：http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?pid=1020&ojid=1&cid=11842&hide=1&problem=Problem%20%20U

原题：

Problem 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

Sample Output

6
0

题意：给出n和n个数，每次只能交换相邻两个元素，求多少次交换后能使这些数升序排列。

思路：典型的求逆序数的问题。由于数组元素可能过大，所以需要用到树状数组离散化。

源代码：

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#define MAX 500010
using namespace std;
int C[MAX];
int b[MAX];//备用数组，用于储存初始数组重新赋值后的结果
int lowbit(int x)
{
    return x &(-x);
}
int getsum(int x)
{
    int res = 0;
    while (x > 0)
    {
        res += C[x];
        x -= lowbit(x);
    }
    return res;
}
void add(int x,int v)
{
    while (x < MAX)
    {
        C[x] += v;
        x += lowbit(x);
    }
}
struct Point//用于储存该点的值及其初始位置
{
    int v;
    int id;
}point[MAX];
bool cmp(Point a,Point b)
{
    return a.v < b.v;
}
int main()
{
    int n;
    while (scanf("%d",&n)!=EOF)
    {
        if (n == 0)
            break;
        for (int i = 1; i <= n; i++)
        {
            scanf("%d",&point[i].v);
            point[i].id = i;
        }
        sort (point+1,point + n + 1,cmp);
        memset(b,0,sizeof(b));
        memset(C,0,sizeof(C));//初始值一定为0
        b[point[1].id] = 1;
        for (int i = 2; i <= n; i++)//数组离散化
        {
            if (point[i].v == point[i-1].v)
                b[point[i].id] = b[point[i-1].id];
            else
                b[point[i].id] = i;
        }
        //for (int i = 1; i <= n; i++)
        //    cout << b[i] << " ";
        //cout << endl;
        long long ans = 0;
        for (int i = 1; i <= n; i++)//求逆序数
        {
            add(b[i],1);
            ans += getsum(n) - getsum(b[i]);
        //    cout << ans << endl;
        }
        //for (int i = 1; i <= n; i++)
        //    cout << C[i] << " ";
        //cout << endl;
            printf("%lld\n",ans);
    }
    return 0;
}