POJ 2299 Ultra-QuickSort （树状数组+离散化）-优快云博客

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Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 63336		Accepted: 23627

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

Source

Waterloo local 2005.02.05

【思路】

显然是要求逆序对的数目了，由于数据范围过大，而数的数目较小，需要离散化后运用树状数组来维护，前缀和函数sum的含义就是不大于某个数的的数的数目，sum(n)-sum(x)就是大于x的数的数目了。

【代码】

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN=500005;

struct point{
    int val,pos;

    bool operator<(const point &another)const
    {
        return val<another.val;
    }
};

int n;
int id[MAXN],num[MAXN];
point a[MAXN];

int lowbit(int x)
{
    return (x&-x);
}

void modify(int x,int add)
{
    while(x<=n){
        num[x]+=add;
        x+=lowbit(x);
    }
}

int sum(int x)
{
    int ans=0;
    while(x>0){
        ans+=num[x];
        x-=lowbit(x);
    }
    return ans;
}

int main()
{
    while(scanf("%d",&n)==1&&n!=0){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i].val);
            a[i].pos=i;
        }
        sort(a+1,a+1+n);
        for(int i=1;i<=n;i++)
            id[a[i].pos]=i;
        memset(num,0,sizeof(num));
        long long ans=0;
        for(int i=1;i<=n;i++){
            modify(id[i],1);
            ans+=(sum(n)-sum(id[i]));
        }
        printf("%lld\n",ans);
    }
    return 0;
}