HDU1542 Line segment tree geometry application scan line

Explain about the scan lines because what is not drawing
a direct reference this blog Figure understand

Note that the nodes in the line segment tree maintain line segments,
so if tree[l] is a leaf node, it does not maintain the interval [l,l] but the line segment [l,l+1], [L, R] is [L,R+1],
so in pushup is lsh[r+1]-lsh[l];
and when the left and right coordinates are updated between the two partitions, because updating 0-2 is maintaining 0-3, so r The position of the node must be -1

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int MAXN = 1e5+7;
struct node
{
    
    
	double l,r,h;
	int f;//l线段左端点 r右端点 h高度 f标记当前这条线段是矩形的上边界还是下边界
}line[MAXN];
double lsh[MAXN];
double tree[MAXN<<2];
int ct[MAXN];
int n;
bool cmp(node a,node b){
    
    
	return a.h < b.h;
}

void pushup(int rt,int l,int r)
{
    
    
	if(ct[rt]) tree[rt] = lsh[r+1]-lsh[l];
	else if(l == r) tree[rt] = 0;
	else tree[rt] = tree[rt<<1]+tree[rt<<1|1];
}

void build(int rt,int l,int r)
{
    
    
	tree[rt] = 0;
	ct[rt] = 0;
	if(l == r) return ;
	int mid = (l+r)>>1;
	build(rt<<1,l,mid);
	build(rt<<1|1,mid+1,r);
}

void update(int rt,int l,int r,int ul,int ur,int x)
{
    
    
	if(l>=ul&&r<=ur){
    
    
		ct[rt] += x;
		pushup(rt,l,r);//覆盖标记发生了变化 因此需要 下推一下 重新复值
		return ;
	}
	int mid = (l+r)>>1;
	if(ul <= mid) update(rt<<1,l,mid,ul,ur,x);
	if(ur > mid) update(rt<<1|1,mid+1,r,ul,ur,x);
	pushup(rt,l,r);
}

int main()
{
    
    
	int cas = 0;

	double x1,x2,y1,y2;
	while(~scanf("%d",&n)&&n){
    
    
		int cnt = 0;
		for(int i = 1;i <= n;i ++){
    
    
			scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
			lsh[++cnt] = x1;//因为坐标会有小数 所以需要离散化 才能用线段树维护
			line[cnt].l = x1;
			line[cnt].r = x2;
			line[cnt].h = y1;
			line[cnt].f = 1;//因为要从下往上扫 所以把下面的当做1
			lsh[++cnt] = x2;
			line[cnt].l = x1;
			line[cnt].r = x2;
			line[cnt].h = y2;
			line[cnt].f = -1;
		}
		sort(lsh+1,lsh+cnt+1);
		sort(line+1,line+cnt+1,cmp);
		int num = unique(lsh+1,lsh+1+cnt)-lsh-1;
		//printf("%d\n",num);
		build(1,1,num);
		double ans = 0;
		for(int i = 1;i <= cnt;i ++){
    
    
			int l = lower_bound(lsh+1,lsh+1+num,line[i].l)-lsh;
			int r = lower_bound(lsh+1,lsh+1+num,line[i].r)-lsh-1;
			//printf("%d %d\n",l,r);
			update(1,1,num,l,r,line[i].f);
			ans += tree[1]*(line[i+1].h-line[i].h);
		}
		printf("Test case #%d\n",++cas);
		printf("Total explored area: %.2f\n",ans);
		printf("\n");
	}
	return 0;
}