https://www.luogu.com.cn/problem/P1502
发现正常扫描线很难维护恰好大小为 W W W 的区间
反过来,对于每个星星维护合法的左下角下标
把原先的判定转成了和点有关的状态,把点变成矩形后求并即可
cpp
#include <algorithm>
#include<iostream>
#include <stdio.h>
using namespace std;
#define int long long
inline 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<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
//mt19937 rand(time(0));
//mt19937_64 rand(time(0));
//srand(time(0));
#define N 20010
//#define M
//#define mo
struct node {
int l, r, x, w;
}a[N];
int n, m, i, j, k, T;
int ans, sum, l, r, x, y, w, Lx, Rx;
int W, H, rt, b[N];
bool cmp(node x, node y) {
if(x.x==y.x) return x.w<y.w;
return x.x<y.x;
}
struct Segment_tree {
int tot, ls[N<<2], rs[N<<2];
int mx[N<<2], tag[N<<2];
void build(int &k, int l, int r) {
if(!k) k=++tot, mx[k]=tag[k]=ls[k]=rs[k]=0;
if(l==r) return ;
int mid=(l+r)>>1;
build(ls[k], l, mid);
build(rs[k], mid+1, r);
}
void push_down(int k) {
tag[ls[k]]+=tag[k]; mx[ls[k]]+=tag[k];
tag[rs[k]]+=tag[k]; mx[rs[k]]+=tag[k];
tag[k]=0;
}
void add(int k, int l, int r, int x, int y, int z) {
if(l>=x && r<=y) return mx[k]+=z, tag[k]+=z, void();
int mid=(l+r)>>1;
push_down(k);
if(x<=mid) add(ls[k], l, mid, x, y, z);
if(y>=mid+1) add(rs[k], mid+1, r, x, y, z);
mx[k]=max(mx[ls[k]], mx[rs[k]]);
}
}Seg;
signed main()
{
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
T=read();
while(T--) {
// while(~scanf("%lld%lld%lld", &n, &W, &H)) {
n=read(); W=read(); H=read();
ans=k=j=0;
for(i=1; i<=n; ++i) {
x=read(); y=read(); w=read();
Lx=x; Rx=x+W-1; //[Lx, Rx]
b[++k]=Lx; b[++k]=Rx;
a[++j].l=Lx; a[j].r=Rx; a[j].x=y; a[j].w=w;
a[++j].l=Lx; a[j].r=Rx; a[j].x=y+H; a[j].w=-w;
}
sort(b+1, b+2*n+1);
for(i=1; i<=2*n; ++i) a[i].l=lower_bound(b+1, b+2*n+1, a[i].l)-b;
for(i=1; i<=2*n; ++i) a[i].r=lower_bound(b+1, b+2*n+1, a[i].r)-b;
Seg.tot=rt=ans=0; Seg.build(rt, 1, 2*n);
sort(a+1, a+2*n+1, cmp);
for(i=1; i<=2*n; ++i) {
Seg.add(1, 1, 2*n, a[i].l, a[i].r, a[i].w);
ans=max(ans, Seg.mx[1]);
}
printf("%lld\n", ans);
}
return 0;
}