【bzoj1071】[SCOI2007]组队

sum= A*h+B*s排序

然后枚举height和speed的最小值

然后用两个指针：先枚举speed最小值，然后一边枚举v的最小值一边查询符合条件的人数。

 

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath> 
using namespace std;
 
typedef long long LL;
 
#define N 10010
 
struct data
{
    LL h,s,sum;
}x[N],y[N];
 
LL n,A,B,C;
LL minn,maxn;
LL ans,cnt,l,r;
 
LL getLL()
{
    LL x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
 
 
LL cmp1(const data &a,const data &b)
{
    return a.h<b.h;
}
 
LL cmp2(const data &a,const data &b)
{
    return a.sum<b.sum;
}
 
bool check1(LL p)
{
    return y[p].s<=maxn && y[p].s>=minn;
}
 
bool check2(LL p)
{
    return x[p].s<=maxn && x[p].s>=minn;
}
 
int main()
{
    n=getLL(),A=getLL(),B=getLL(),C=getLL();
    for (LL i=1;i<=n;i++)
    {
        x[i].h=getLL();
        x[i].s=getLL();
        x[i].sum=A*x[i].h+B*x[i].s;
        y[i]=x[i];
 
    }
    sort(x+1,x+n+1,cmp1);
    sort(y+1,y+n+1,cmp2);
    for (LL i=1;i<=n;i++)
    {
        minn=x[i].s;
        maxn=minn+C/B;
        l=r=cnt=0;
        for (LL j=1;j<=n;j++)
        {
            while (r<n && y[r+1].sum<=A*x[j].h+B*x[i].s+C)
                r++,cnt+=check1(r);
            while (l<n && x[l+1].h<x[j].h)
                l++,cnt-=check2(l);
            ans=max(ans,cnt);
        }
    }
    printf("%lld\n",ans);
    return 0;
}