【SGU 180】Inversions —— 归并排序或树形数组计算逆序对

原题链接

180. Inversions

time limit per test: 0.25 sec.

memory limit per test: 4096 KB

input: standard

output: standard

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

3

解题报告：

逆序对！经典的算法之一！

方法一是利用归并排序来计算，简单修改merge即可统计逆序对的数量。

方法二是采用离散化+树形数组（或者线段树）

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;

long long reverse_pair(int a[],int n)
{
    if(n<2) return 0;

    long long sum=0 , mid=n/2;
    sum += reverse_pair(a,mid);
    sum += reverse_pair(a+mid,n-mid);

    int *b=new int[n];
    memcpy(b,a,n*sizeof(int));

    int i=0,j=mid; int k=0;
    for(;i<mid && j<n;++k)
        if(b[i]<=b[j]) {a[k]=b[i];++i;}
        else {a[k]=b[j];++j;sum+=(mid-i);}

    if(j==n) for(;i<mid;++i,++k) a[k]=b[i];
    else     for(;j<n;++j,++k) a[k]=b[j];

    delete [] b;
    return sum;
}

int C[65537+10],n;
inline int lowbit(int x){return x&(-x);}
long long sum(int k){
    long long sum=0;
    while(k>0) {
        sum+=C[k]; k-=lowbit(k);
    }
    k=n;
    while(k>0) {
        sum-=C[k]; k-=lowbit(k);
    }
    return -sum;
}
void modify(int k){
    while(k<=n){C[k]+=1;k+=lowbit(k);}
}

int main()
{
     int a[65537+10],b[65537+10];
     while(cin>>n){
//        方法1：归并排序
//        for(int i=0;i<n;++i)
//            cin>>a[i];
//        cout<<reverse_pair(a,n)<<endl;

//        方法2：离散化+树形数组
        for(int i=0;i<n;++i)  cin>>a[i];
        memcpy(b,a,n*sizeof(int));

        // 离散化
        sort(a,a+n);
        int SIZE=unique(a,a+n)-a;
        for(int i=0;i<n;++i) b[i]=lower_bound(a,a+SIZE,b[i])-a+1;

        // 树形数组
        long long ans=0; memset(C,0,sizeof(C));
        for(int i=0;i<n;++i){
            modify(b[i]);
            ans+=sum(b[i]);
        }
        cout<<ans<<endl;
     }
}