POJ 2299 Ultra-QuickSort（树状数组）

朱雀 2022-06-16 06:55 169阅读 0赞

> Ultra-QuickSort  
> Time Limit: 7000MS Memory Limit: 65536K  
> Total Submissions: 60584 Accepted: 22464  
> 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

> 题解：求逆序对数，可以用归并排序，树状数组实现（离散化：开一个大小为这些数的最大值的树状数组）

> 代码：

#include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    
    typedef long long ll;
    using namespace std;
    const int maxn = 500005;
    int n;
    int c[maxn],a[maxn],b[maxn];
    //val就是原输入的值，pos是下标
    // 对结构体按val从小到大排序
    //此时，val和结构体的下标就是一个一一对应关系
    struct Node
    {
        int val;
        int pos;
        bool operator <(const Node &a)
        {
            return val<a.val;
        }
    }node[maxn];
    
    int lowbit(int x)
    {
        return x&(-x);
    }
    int get_sum(int x)
    {
        int sum=0;
        while(x)
        {
            sum+=c[x];
            x -=lowbit(x);
        }
        return sum;
    }
    void add(int i,int x)
    {
        while(i<=n)
        {
            c[i] +=x;
            i +=lowbit(i);
        }
    }
    
    int main()
    {
        while(cin>>n&&n)
        {
            memset(a,0,sizeof(a));
            memset(c,0,sizeof(c));
            for(int i=1;i<=n;i++)
            {
                cin>>node[i].val;
                node[i].pos=i;
            }
            sort(node+1,node+1+n);
            for(int i=1;i<=n;i++)
                a[node[i].pos]=i;///离散化
    
            ll ans=0;
            for(int i=1;i<=n;i++)
            {
                ans +=i -1-get_sum(a[i]);//统计逆序数
                add(a[i],1);
            }
            cout<<ans<<endl;
        }
        return 0;
    }