[] Window Luogu P1502 stars

Luogu P1502

Well understood meaning of the questions is asked rectangle given the maximum and trap.

But doing it is very troublesome.

- crazy burst of anger from 10 points

The idea is a more efficient - a star to each bottom left to the upper right as a rectangle is formed to expand,
expansion rules as long as the top right corner of the window within the rectangles, can cover the stars. Then maintaining a scanning line segment tree from left to right sweep in the past, to find the maximum value of a single point.

It is noteworthy that the question surfaces made of stars on the window frame are not included in the answer, so do a whole questions becomes very sick

Better understanding and a more convenient approach is to horizontal and vertical coordinates +0.5 stars as the lower left corner of the rectangle.

This will then ensure compliance with this rule to expand the rectangle.

In addition, due to the large coordinate the stars, you can use discrete narrow range

code time:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define lson root<<1
#define rson root<<1|1
#define ll long long
using namespace std;
struct data
{
    double l,r,x;
    ll flag;
}line[80005];
bool cmp(data a,data b)
{
    if (a.x==b.x) return a.flag<b.flag;
    return a.x<b.x;
    //重点之一，注意权值小的排在前面，因为在矩形的右边界上，这颗星星已经对答案没有贡献了
}
ll tag[80005],tree[80005],n,w,h,x,y,l,cnt,ans,T;
double pnt[80005];
void push_down(ll root,ll l,ll r)
{
    tag[lson]+=tag[root];
    tag[rson]+=tag[root];
    tree[lson]+=tag[root];
    tree[rson]+=tag[root];
    tag[root]=0;    
}*/
void update(ll root,ll l,ll r,double L,double R,ll flag)
{
    if (L<=pnt[l]&&pnt[r]<=R)
    {
        tag[root]+=flag;
        tree[root]+=flag;
        return ;
    }
    if (l+1==r) return ;
    if (R<=pnt[l]||L>=pnt[r]) return ;
    push_down(root,l,r);
    //事实上不需要pd操作也能过。
    ll mid=(l+r)>>1;
    if (L<pnt[mid]) update(lson,l,mid,L,R,flag);
    if (R>pnt[mid]) update(rson,mid,r,L,R,flag);
    //注意离散化后mid仅为下标，而不是坐标。
    tree[root]=max(tree[lson],tree[rson])+tag[root];
}
int main()
{
    scanf("%d",&T);
    for (int q=1;q<=T;q++)
    {
        cnt=0;
        ans=0;
        memset(line,0,sizeof(line));
        memset(pnt,0,sizeof(pnt));
        memset(tree,0,sizeof(tree));
        memset(tag,0,sizeof(tag));
        //记得要初始化
        scanf("%d%d%d",&n,&w,&h);
        for (int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&x,&y,&l);
            line[++cnt].x=x+0.5;line[cnt].l=y+0.5;line[cnt].r=y+h;line[cnt].flag=l;pnt[cnt]=y+h;
            line[++cnt].x=x+w;line[cnt].l=y+0.5;line[cnt].r=y+h;line[cnt].flag=-l;pnt[cnt]=y+0.5;
            //重点之一,对边界的处理
        }
        sort(line+1,line+1+cnt,cmp);
        sort(pnt+1,pnt+1+cnt);
        ll til=unique(pnt+1,pnt+1+cnt)-pnt-1;
        for (int i=1;i<=cnt;i++)
        {
            update(1,1ll,til,line[i].l,line[i].r,line[i].flag);
            ans=max(ans,tree[1]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}