SGU180 Inversions 离散 + 树状数组

原创于 2018-07-20 12:58:27 发布 · 202 阅读

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本文介绍了一种利用树状数组进行逆序对计数的方法，适用于离散化的整数集合。通过输入一系列整数，该算法能高效计算出所有逆序对的数量。

There are N integers (1<=N<=65537) A1, A2,.. AN (0<=Ai<=10^9). You need to find amount of such pairs (i, j) that 1<=i<j<=N and A[i]>A[j].
Input

The first line of the input contains the number N. The second line contains N numbers A1...AN.
Output

Write amount of such pairs.
Sample test(s)
Input

5
2 3 1 5 4
Output

题解：离散化之后树状数组统计求和即可。

AC代码：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define _for(i,a,b) for(int i=a;i<=b;i++)
typedef long long ll;
const int maxn = 70000;
struct node
{
    int id,num;
    bool operator<(const node& a)const
    {
        return num<a.num;
    }
}a[maxn];
int n,b[maxn];
ll c[maxn];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x)
{
    while(x<=n)
    {
        c[x]++;
        x+=lowbit(x);
    }
}
ll sum(int x)
{
    ll now = 0;
    while(x>0)
    {
        now+=c[x];
        x-=lowbit(x);
    }
    return now;
}
int main(int argc, char const *argv[])
{
    cin>>n;
    _for(i,1,n)
    {
        cin>>a[i].num;
        a[i].id=i;
   }
    sort(a+1,a+1+n);
    b[a[1].id]=1;
    _for(i,2,n)
    {
        if(a[i].num!=a[i-1].num)b[a[i].id]=i;
        else b[a[i].id]=b[a[i-1].id];
    }
    ll ans = 0;
    _for(i,1,n)
    {
        update(b[i]);
        ans+=sum(n)-sum(b[i]);
    }
    cout<<ans<<endl;        
    return 0;
}