求助,归并排序!!!全WA

P1908 逆序对

tzl_Dedicatus545 @ 2020-04-20 12:09:03

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

long long a[600005],tmp[600005],cnt;

void merge(long long left,long long mid,long long right)
{
    long long i=left,j=mid+1,k=left;

    while(i<=mid && j<=right)
    {
        if(a[i]<=a[j])
        {
            tmp[k++]=a[i++];
        }
        else
        {
            cnt++;
            tmp[k++]=a[j++];
        }
    }

    while(i<=mid)   
    {
        tmp[k++]=a[i++];
    }
    while(j<=right)
    {
        tmp[k++]=a[j++];
    }

    for(long long i=left;i<=right;i++)
    {
        a[i]=tmp[i];
    }
}

void mergesort(long long left,long long right)
{   
    if(left>=right)
        return ;
    long long mid=(left+right)/2;

    mergesort(left,mid);
    mergesort(mid+1,right);

    merge(left,mid,right);
}

int main()
{
    ios::sync_with_stdio(false); \\快读
    long long n;

    cin>>n;

    for(long long i=1;i<=n;i++)
        cin>>a[i];

    mergesort(1,n);

    cout<<cnt;

    return 0;
}

评测记录


by PragmaGCC @ 2020-04-20 12:10:14

您归并排序假的?


by FZzzz @ 2020-04-20 12:11:09

@ljm1069 不能 cnt++ 啊,因为它会对很多元素都造成贡献


by PragmaGCC @ 2020-04-20 12:11:14


cnt++
改成
cnt += mid-i+1

by tzl_Dedicatus545 @ 2020-04-20 12:47:10

@反面教材945 请问能说的详细点吗


by tzl_Dedicatus545 @ 2020-04-20 12:48:31

@反面教材945 不太懂


by WanderOvO @ 2020-04-20 13:35:29

数列分为了左半段和右半段,现在,我们发现左半段的a[i]大于右半段的a[j],这说明在左半段中,a[i]以及他后面的元素都比a[j]大,都会和a[j]构成逆序对。所以cnt应该按照楼上说的写


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