The Number of Inversions（逆序数）

标签：version   gray   length   one   math   display   false   char   roc   

For a given sequence A={a0,a1,...an−1}A={a0,a1,...an−1}, the number of pairs (i,j)(i,j) where ai>ajai>aj and i<ji<j, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program:

bubbleSort(A)
  cnt = 0 // the number of inversions
  for i = 0 to A.length-1
    for j = A.length-1 downto i+1
      if A[j] < A[j-1]
	swap(A[j], A[j-1])
	cnt++

  return cnt

For the given sequence AA, print the number of inversions of AA. Note that you should not use the above program, which brings Time Limit Exceeded.

Input

In the first line, an integer nn, the number of elements in AA, is given. In the second line, the elements aiai (i=0,1,..n−1i=0,1,..n−1) are given separated by space characters.

output

Print the number of inversions in a line.

Constraints

• 1≤n≤200,0001≤n≤200,000
• 0≤ai≤1090≤ai≤109
• aiai are all different

Sample Input 1

5
3 5 2 1 4

Sample Output 1

6

Sample Input 2

3
3 1 2

Sample Output 2

2

已知逆序数等于冒泡排序的序列，但这题冒泡排序肯定超时。这题用归并排序优化一下就行。

AC代码

#include<iostream>
#include<cstring>
#include<stack>
#include<cstdio>
#include<cmath>
using namespace std;
#define MAX 500000
#define INF 2e9
int L[MAX/2+2],R[MAX/2+2];
long long  cnt=0;
long long merge(int A[],int n,int left,int mid,int right)
{
    long long cnt=0;
    int n1=mid-left;
    int n2=right-mid;
    for(int i=0;i<n1;i++)
    {
        L[i]=A[left+i];
    }
    for(int i=0;i<n2;i++)
    {
        R[i]=A[mid+i];
    }
    L[n1]=INF;
    R[n2]=INF;
    int i=0,j=0;
    for(int k=left;k<right;k++)//合并
    {
     if(L[i]<=R[j])
     A[k]=L[i++];
     else
     {
     A[k]=R[j++];
     cnt=cnt+(n1-i);
}
}
return cnt;
}
long long  mergeSort(int A[],int n,int left,int right)
{
    long long v1,v2,v3;
    if(left+1<right)
    {
        int mid=(left+right)/2;
        v1=mergeSort(A,n,left,mid);
        v2=mergeSort(A,n,mid,right);
        v3=merge(A,n,left,mid,right); 
        return (v1+v2+v3); 
    }
    else
    return 0;
}
int main()
{
int A[MAX],n;
cnt=0;
cin>>n;
for(int i=0;i<n;i++)
cin>>A[i];
cnt=mergeSort(A,n,0,n);
cout<<cnt<<endl;
return 0;
 }

The Number of Inversions（逆序数）

标签：version   gray   length   one   math   display   false   char   roc   

原文地址：https://www.cnblogs.com/hh13579/p/10859434.html