树状数组求逆序数 POJ 2299

转自：https://blog.csdn.net/qq_41105401/article/details/79886166

Ultra-QuickSort

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：这题是经典的树状数组求逆序数。由于数比较大先离散化再用树状数组求。

用树状数组求逆序数时候初学者可能有些晕乎，而其他某些博客又有些错误或不易理解，这里就来讲解一下。

首先设树状数组要维护的数组是c[ ],先将其初始化为0

对于每个位置i上的数a[i]，它原本应该在的位置是u = a[i]，那么就让c[u] = 1，那么i之前小于a[i]的数的个数就是sum[u] - 1，

那么i之前大于a[i]的数的个数就是 i - 1 - （sum[u]-1） =i - sum[u] 

代码：

时间复杂度O(n*logn)

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int rem[500005];
int c[500005];
int n;
int Lowbit(int x)  // 2^k
{
    return x & (-x);
}
void update(int i, int x)//i点增量为x
{
    while(i <= n)
    {
        c[i] += x;
        i += Lowbit(i);
    }
}
int sum(int x)//区间求和 [1,x]
{
    int sum = 0;
    while(x > 0)
    {
        sum += c[x];
        x -= Lowbit(x);
    }
    return sum;
}
struct node
{
    int value;
    int before;
}a[500005];
int mmp[500005];
bool compare(node x,node y)
{
    return x.value < y.value;
}
int main()
{
    while(~scanf("%d",&n))
    {
        long long res = 0;
        memset(c,0,sizeof(c));
        if(!n)
            break;
        for(int i = 1; i <= n ; ++i)
        {
            scanf("%d",&a[i].value);
            a[i].before = i;
        }
        sort(a+1,a+n+1,compare);
/*
这段离散化做别的题时候发现有错误
        for(int i = 1; i <= n; ++i)
        {
            rem[a[i].before] = i;
        }
*/
//改成这个了：
        int cnt = 1;
        for(int i=1;i<=n;i++){
            if(i != 1 && a[i].value != a[i-1].value)
                cnt++;
            rem[a[i].before] = cnt;
        }
        
        
        for(int i = 1; i <= n; ++i)
        {
            update(rem[i],1);
            res += i-sum(rem[i]);
        }
        cout<<res<<endl;
    }
    return 0;
}

flyzer的代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<vector>
#include<map>
#include<cstdlib>
#define ll long long
#define inf 0x3f3f3f3f//1e9+6e7
#define N 500010
#define debug(x) cout<<"X: "<<(x)<<endl
#define de cout<<"**"<<endl
#define lowbit(x) ((x)&(-x))
const double pi=acos(-1.0);
using namespace std;
int n;
int a[N],b[N],c[N];
void update(int i,int val)
{
    while(i<=n)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int ret=0;
    while(i>0)
    {
        ret+=c[i];
        i-=lowbit(i);
    }
    return ret;
}
int main()
{
    while(scanf("%d",&n)&&n)
    {
        memset(a,0,sizeof a);
        memset(c,0,sizeof c);
        //**************************
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            b[i]=a[i];
        }
        sort(b,b+n);
        for(int i=0;i<n;i++)
            a[i]=(lower_bound(b,b+n,a[i])-b)+1; //因为树状数组不能处理下标从0开始的，所以这里要加1
        //*********************离散化
        ll ans=0;
        for(int i=0;i<n;i++)
        {
            ans+=(i-sum(a[i]));
            update(a[i],1);
        }
        cout<<ans<<endl;
    }
    return 0;
}

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