JZOJ 数数

Posted 2020-10-18

二进制通常通过拆位来做，类欧几里得f裸题，直接上代码；

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

const int N=2505;
const int inf=1000000000;
const int M=2500;

int n,s[N][N],ur[N][N],dl[N][N],ul[N][N],dr[N][N],drd[N][N],drr[N][N],ulu[N][N],ull[N][N];
struct data{int x,y;}a[250005];

int read()
{
    int x=0,f=1;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while (ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}

int get_dl(int x,int y)
{
    if (x<1||x>M||y<1||y>M) return 0;
    if (dl[x][y]>-1) return dl[x][y];
    dl[x][y]=0;
    int d=min(drd[x][y+1],x),l=min(ull[x+1][y],y);
    return dl[x][y]=get_dl(d,l)+dr[x][1]-dr[x][y+1];
}

int get_ur(int x,int y)
{
    if (x<1||x>M||y<1||y>M) return 0;
    if (ur[x][y]>-1) return ur[x][y];
    ur[x][y]=0;
    int u=max(ulu[x][y-1],x),r=max(drr[x-1][y],y);
    return ur[x][y]=get_ur(u,r)+ul[x][M]-ul[x][y-1];
}

int main()
{
    n=read();
    for (int i=1;i<=n;i++) a[i].x=read(),a[i].y=read(),a[i].y=M-a[i].y+1,s[a[i].x][a[i].y]++;
    for (int i=0;i<=M+1;i++)
        for (int j=0;j<=M+1;j++)
        {
            ull[i][j]=drd[i][j]=inf;
            ulu[i][j]=drr[i][j]=-1;
            dl[i][j]=ur[i][j]=-1;
        }
    for (int i=M;i>=1;i--)
        for (int j=1;j<=M;j++)
        {
            ull[i][j]=min(ull[i][j-1],ull[i+1][j]);
            ulu[i][j]=max(ulu[i][j-1],ulu[i+1][j]);
            if (s[i][j]) ull[i][j]=min(ull[i][j],j),ulu[i][j]=max(ulu[i][j],i);
            ul[i][j]=ul[i][j-1]+ul[i+1][j]-ul[i+1][j-1]+s[i][j];
        }
    for (int i=1;i<=M;i++)
        for (int j=M;j>=1;j--)
        {
            drr[i][j]=max(drr[i][j+1],drr[i-1][j]);
            drd[i][j]=min(drd[i][j+1],drd[i-1][j]);
            if (s[i][j]) drr[i][j]=max(drr[i][j],j),drd[i][j]=min(drd[i][j],i);
            dr[i][j]=dr[i][j+1]+dr[i-1][j]-dr[i-1][j+1]+s[i][j];
        }
    for (int i=1;i<=n;i++)
    {
        int x=a[i].x,y=a[i].y;
        int ans=get_dl(x,y)+get_ur(x,y)+n-3;
        printf("%d\n",ans);
    }
    return 0;
}