POJ-2299 Ultra_QuickSort 线段树+逆序对数

最新推荐文章于 2020-01-13 16:07:47 发布

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最新推荐文章于 2020-01-13 16:07:47 发布

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原文链接：http://www.cnblogs.com/DaD3zZ-Beyonder/p/5346254.html

本文介绍了一种使用线段树求解逆序对的方法，通过离散化处理大数据范围，实现对给定序列进行冒泡排序交换次数的高效计算。

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Ultra-QuickSort
Time Limit: 7000MS Memory Limit: 65536K
Total Submissions: 50737 Accepted: 18595

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

Source
Waterloo local 2005.02.05

题目大意：给定一个数列，求冒泡排序交换次数（逆序对数）

线段树求逆序对，在数据范围过大的时候，需要进行离散化，然后进行建树，基本题

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 500002
int sum[maxn<<2]={0};
struct data{
    int num,loc;
};
int f2[maxn]={0};
data f1[maxn]={0};

int cmp(data x,data y)
{
    return x.num<y.num;
}

void updata(int now)
{
    sum[now]=sum[now<<1]+sum[now<<1|1];
}

void point_change(int now,int l,int r,int loc)
{
    if (l==r)
        {
            sum[now]++;
            return;
        }
    int mid=(l+r)>>1;
    if (loc<=mid)
        point_change(now<<1,l,mid,loc);
    else
        point_change(now<<1|1,mid+1,r,loc);
    updata(now);
}

int query(int L,int R,int l,int r,int now)
{
    if (L<=l && R>=r)
        return sum[now];
    int mid=(l+r)>>1;
    int ans=0;
    if (L<=mid)
        ans+=query(L,R,l,mid,now<<1);
    if (R>mid)
        ans+=query(L,R,mid+1,r,now<<1|1);
    return ans;
}

int main()
{
    int n;
    while (true)
        {
            scanf("%d",&n);
            memset(f1,0,sizeof(f1));
            memset(f2,0,sizeof(f2));
            memset(sum,0,sizeof(sum));
            if (n==0) break;
            long long number=0;
            for (int i=1; i<=n; i++)
                    {
                        scanf("%d",&f1[i].num);
                        f1[i].loc=i;
                    }
            sort(f1+1,f1+n+1,cmp);
            for (int i=1; i<=n; i++)
                f2[f1[i].loc]=i;//离散化部分(感觉这个离散化写的巨不专业） 
            //for (int i=1; i<=n; i++)
                //cout<<f2[i]<<' ';
            //cout<<endl;
            for (int i=1; i<=n; i++)
                {
                    int x=f2[i];
                    number+=query(x,n,1,n,1);
                    point_change(1,1,n,x);
                }//逆序对的求法，每次从这个数到n求和，找出比他大的且出现在它之前的 
            printf("%d",number);
        }
    return 0;
}