Reverse problem P1309, P1908, P1966

Switzerland round, merge sort simplified time

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// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
struct node{int w,st,id;}num[100001*2],a[100001*2],b[100001*2];
int n,r,q;
int cmp(node x,node y){if(x.w==y.w)return x.id<y.id;else  return x.w>y.w;}
void gb()
{
    int p1=1,p2=1;
    for(int i=1;i<=n*2;i++)
    if(a[p1].w>b[p2].w||(a[p1].w==b[p2].w&&a[p1].id<b[p2].id))
    num[i]=a[p1++];
    else num[i]=b[p2++];
}
int main ()
{
    cin>>n>>r>>q;
    for(int i=1;i<=n*2;i++)cin>>num[i].w;
    for(int i=1;i<=n*2;i++)cin>>num[i].st;
    for(int i=1;i<=n*2;i++)num[i].id=i;
    sort(num+1,num+1+2*n,cmp);
    while(r--)
    {
    
        for(int i=1;i<=2*n;i=i+2)
        if(num[i].st>num[i+1].st)
        ne [i / 2 + 1 ] = num [i], b [i / 2 + 1 ] = num [i + 1 ], a [i / 2 + 1 ] .w ++ ;}
         Else  
        {a [i / 2 + 1 ] = num [i + 1 ], b [i / 2 + 1 ] = num [i] and [i / 2 + 1 ] .w ++ ;}    
        gb();
    }
    cout<<num[q].id;
}

Reverse order, the use of merge sort Statistics

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#include<bits/stdc++.h>
using namespace std;
#define ll long long 
int n,a[500010],b[500010];
ll ans;
int read()
{
    int fg=1;ll cn=0;char ch=getchar();
    while((ch<'0'||ch>'9')){if(ch=='-')fg=-1;ch=getchar();}
    while((ch>='0'&&ch<='9'))cn=cn*10+ch-'0',ch=getchar();
    return cn*fg;
}
void msort(int l,int r)
{
    if(l==r)return;
    int mid=(l+r)/2;
    msort(l,mid);msort(mid+1,r);
    int p1=l,p2=mid+1,k=l;
    while(p1<=mid&&p2<=r)
        if(a[p1]<=a[p2])b[k++]=a[p1++];
        else b[k++]=a[p2++],ans+=(ll)(mid-p1+1);
    while(p1<=mid)b[k++]=a[p1++];
    while(p2<=r)b[k++]=a[p2++];
    for(int i=l;i<=r;i++)a[i]=b[i];
}
int main () {
    cin>>n;
    for(int i=1;i<=n;i++)
    a[i]=read();
    msort ( 1 , n);
    cout << years;
}

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Match line up, first use the sort inequality come to need discrete, then reverse after solving discrete

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