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  • POJ-2299-Ultra-QuickSort(单点更新 + 区间查询+离散化)

    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
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    #include<stack>
    #include<set>
    #include<vector>
    #include<map>
    const int maxn=5e5+5;
    typedef long long ll;
    using namespace std;
    struct node
    {
        ll l,r;
        ll sum;
    }tree[maxn<<2];
    int a[maxn],sub[maxn],cnt;
    
    void pushup(int m)
    {
        tree[m].sum=(tree[m<<1].sum+tree[m<<1|1].sum);
    }
    //void pushdown(int  m)
    //{
    //    if(tree[m].lazy)
    //    {
    //        tree[m<<1].lazy=tree[m].lazy;
    //        tree[m<<1|1].lazy=tree[m].lazy;
    //        tree[m<<1].sum=tree[m].lazy*(tree[m<<1].r-tree[m<<1].l+1);
    //        tree[m<<1|1].sum=tree[m].lazy*(tree[m<<1|1].r-tree[m<<1|1].l+1);
    //        tree[m].lazy=0; 
    //    } 
    //    return ;
    //}
    void build(int m,int l,int r)
    {
        tree[m].l=l;
        tree[m].r=r;
        tree[m].sum=0;
        if(l==r)
        {
            tree[m].sum=0;
            return ;
        }
        int mid=(tree[m].l+tree[m].r)>>1;
        build(m<<1,l,mid);
        build(m<<1|1,mid+1,r);
        pushup(m);
        return ;
    }
    void update(int m,int index ,int  val)
    {
        if(tree[m].l==index&&tree[m].l==tree[m].r)
        {
            tree[m].sum++;
            return ;
        }
        int mid=(tree[m].l+tree[m].r)>>1;
        if(index<=mid)
        {
            update(m<<1,index,val);
        }
        else 
        {
            update(m<<1|1,index,val);
        }
        pushup(m);
        return ; 
    }
    ll query(int m,int l,int r)
    {
        if(l>r)
        {
            return 0;
        }
        if(tree[m].l==l&&tree[m].r==r)
        {
            return tree[m].sum;
        } 
    //    pushdown(m);
        int mid=(tree[m].l+tree[m].r)>>1;
        if(r<=mid)
        {
            return query(m<<1,l,r);
        } 
        else if(l>mid)
        {
            return query(m<<1|1,l,r);
        }
        else
        {
            return query(m<<1,l,mid)+query(m<<1|1,mid+1,r);
        }
    }
    
    int main()
    { 
       int n;
       while(cin>>n)
       {
           if(n==0)
           {
               break;
        }
        for(int i=0;i<n;++i)scanf("%d",&sub[i]),a[i]=sub[i];
        sort(sub,sub+n);    
        int size=unique(sub,sub+n)-sub;   
        for(int i=0;i<n;i++)
        a[i]=lower_bound(sub,sub+size,a[i])-sub+1;
        build(1,1,size);
        ll sum=0;
        for(int t=0;t<n;t++)
        {
            sum+=query(1,a[t]+1,size);
            update(1,a[t],1);
        }
        printf("%lld
    ",sum);
        
       }
       return 0;
    }



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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/11298857.html
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