Count Inversion题解

2024年6月9日 213次阅读

Problem Description

Recall the problem of finding the number of inversions. As in the course, we are given a sequence of nn numbers a1,a2,⋯,an and we define an inversion to be a pair i<j such that ai>aj

We motivated the problem of counting inversions as a good measure of how different two orderings are. However, one might feel that this measure is too sensitive. Let’s call a pair a significant inversion if i<j and ai>3aj. Give an O(nlog⁡n) algorithm to count the number of significant inversions between two orderings.

The array contains N elements (1<=N<=100,000). All elements are in the range from 1 to 1,000,000,000.

Input

The first line contains one integer N, indicating the size of the array. The second line contains N elements in the array.

  • 50% test cases guarantee that N < 1000.

Output

Output a single integer which is the number of pairs of significant inversions.

Sample Inout

6
13 8 5 3 2 1

 

Sample Output

6

 题意大题为：求当下标i < j时，满足ai>3aj条件的数值对个数。主要是使用分治算法，首先把整体拆分成两个子问题，直到不能再进行拆分，然后解决一个一个子问题。

#include<iostream>
#include<stdlib.h>
using namespace std;
int count = 0;
int min(int a, int b) {
    return a < b ? a : b;
}
long long int merge(int a[], int left, int mid, int right,int b[]) {
   int i = mid;
   int j = right;
   long long int lcount = 0;
   while (i >= left && j > mid) {
       if(a[i] > (long long) 3 * a[j]) { //此处很有可能造成int已经无法满足要求，因此需要定义为longlongint类型，这里是强制类型转化
           lcount += j - mid;
           i--;
       }else{
           j--;
       }
   }
   i = mid;
   j = right;
   int k = 0;
   while (i >= left && j > mid) {
       if(a[i] > a[j]) {
           b[k++] = a[i--];
       }else{
           b[k++] = a[j--];
       }
   }
   while (i >= left) {
       b[k++] = a[i--];
   }
   while (j > mid) {
       b[k++] = a[j--];
   }
   for (i = 0; i < k; i++) {
       a[right - i] = b[i];
   }
    return lcount;
}

long long int solve(int a[],int left, int right,int b[]) {
    long long int cnt = 0;
    if(right > left){
        int mid = (right+left) / 2;
        cnt += solve(a,left, mid,b);
        cnt += solve(a,mid + 1, right,b);
        cnt += merge(a,left, mid, right,b);
    }
    return cnt;
}
long long int InversePairs(int a[],int len)
{
    int *b=new int[len];
    long long int count=solve(a,0,len-1,b);
    delete[] b;
    return count;
}
int main() {
    int n;
    int i;
    int a[100005];
    cin>>n;
    for(i = 0; i < n; i ++) {
        cin >> a[i];
    }
    long long int count = InversePairs(a, n);
    cout<<count<<endl;
}

 

 

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