洛谷p1502窗口的星星 扫描线

题目链接：https://www.luogu.org/problem/P1502

扫描线的板子题，把每个点看成矩形，存下边(x,y,y+h-1,li)和(x+w-1,y,y+h-1)，在按横坐标扫线段更新y区间，线段树维护区间最大值

#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1
#define maxn 20005
struct seg{
    ll x,l,r,s,val;
    seg(){}
    seg(ll x,ll l,ll r,ll s,ll val):x(x),l(l),r(r),s(s),val(val){};
    bool operator <(const seg &w)const{
        if(x==w.x)return s>w.s;
        return x<w.x;
    }
}se[maxn<<];
struct node{
    ll x,y,val,x1,y1;
}a[maxn<<];
ll n,w,h,ly[maxn<<],ny;
ll sum[maxn<<],lazy[maxn<<];
void pushup(int rt)
{
    sum[rt]=max(sum[rt<<],sum[rt<<|]);
}
void build(int l,int r,int rt)
{
    sum[rt]=lazy[rt]=;
    if(l==r)
    {
        return ;
    }
    int mid=l+r>>;
    build(ls);build(rs);
    pushup(rt);
}
void pushdown(int rt)
{
    if(lazy[rt])
    {
        sum[rt<<]+=lazy[rt];
        sum[rt<<|]+=lazy[rt];
        lazy[rt<<]+=lazy[rt];
        lazy[rt<<|]+=lazy[rt];
        lazy[rt]=;
    }
}
void update(int L,int R,int val,int l,int r,int rt)
{
    if(L<=l&&R>=r)
    {
        sum[rt]+=val;
        lazy[rt]+=val;
        return ;
    }
    pushdown(rt);
    int mid=l+r>>;
    if(L<=mid)update(L,R,val,ls);
    if(R>mid)update(L,R,val,rs);
    pushup(rt);
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n>>w>>h;
        int cnt=;
        for(int i=;i<=n;i++)
        {
            cin>>a[i].x>>a[i].y>>a[i].val;
            a[i].y1=a[i].y+h-;
            ly[++cnt]=a[i].y;ly[++cnt]=a[i].y+h-;
            se[cnt-]=seg(a[i].x,a[i].y,a[i].y1,,a[i].val);
            se[cnt]=seg(a[i].x+w-,a[i].y,a[i].y1,-,a[i].val);
        }
        sort(ly+,ly++cnt);
        sort(se+,se++cnt);
        ny=unique(ly+,ly++cnt)-ly-;
        build(,ny,);
        int l,r;
        ll ans=;
        for(int i=;i<=cnt;i++)
        {
            l=lower_bound(ly+,ly++ny,se[i].l)-ly;
            r=lower_bound(ly+,ly++ny,se[i].r)-ly;
            if(se[i].s==)update(l,r,se[i].val,,ny,);
            else update(l,r,-se[i].val,,ny,);
            ans=max(ans,sum[]);
        }
        cout<<ans<<endl;
    }
    return ;
}