POJ 1151 Atlantis (scan line method, line segment tree, discretization)

POJ 1151 Atlantis (scan line method, line segment tree, discretization)

Blue Book Line Segment Tree Exercises POJ 1151 Atlantis
Question Meaning: Given two coordinate points (x, y) (x, y) at the bottom left and top right of n rectangles$(x,y)$ , find their area and (total area)
data range:$n:100、x/y：1e5$

Idea: I'm so sleepy now and write tomorrow^^

LL n;
struct SegmentTree{
    
    
	int l,r,add;
	double dat;
	#define l(x) tree[x].l
    #define r(x) tree[x].r
    #define dat(x) tree[x].dat
    #define add(x) tree[x].add
}tree[maxn*8];
struct node{
    
    
    double x,y,z;
    int c,zz,yy;
    bool operator <(const node aa)const{
    
    
        return x<aa.x;
    }
}a[maxn*2];
double  b[maxn*2];

void build(int p,int l,int r){
    
    
	l(p)=l,r(p)=r;
	dat(p)=0;add(p)=0;
	if(l==r) {
    
    return;}
	int mid=(l+r)/2;
	build(p*2,l,mid);
	build(p*2+1,mid+1,r);
}

void lazy(int p){
    
    
	if(add(p)){
    
    
		dat(p*2)+=add(p);
		dat(p*2+1)+=add(p);
		add(p*2)+=add(p);
		add(p*2+1)+=add(p);
		add(p)=0;
	}
}

void change(int p,int l,int r,double z){
    
    
	if(l<=l(p)&&r>=r(p)){
    
    dat(p)=((add(p)+=z)?b[r(p)+1]-b[l(p)]:0);}
	if(l(p)==r(p)){
    
    return;}
	int mid=(l(p)+r(p))/2;
	if(l<=mid) change(p*2,l,r,z);
    if(r>mid) change(p*2+1,l,r,z);
	if(add(p))dat(p)=b[r(p)+1]-b[l(p)];
	else dat(p)=dat(2*p)+dat(2*p+1);
}

LL ask(int p,int l,int r)
{
    
    
	if(l<=l(p)&&r>=r(p)) return dat(p);
	lazy(p);
    int mid=(l(p)+r(p))/2;
	LL ans=0;
    if(l<=mid) ans+=ask(p*2,l,r);
    if(r>mid) ans+=ask(p*2+1,l,r);
    return ans;
}

int main(){
    
    
    LL co,sum=0;
    double ans;
    while(scanf("%lld",&n)!=EOF&&n){
    
    
        ans=0;
        for(int i=1;i<=n;i++){
    
    
            scanf("%lf%lf%lf%lf",&a[2*i-1].x,&a[2*i-1].y,&a[i*2].x,&a[2*i].z);
            b[i*2-1]=a[2*i].y=a[2*i-1].y;
            b[2*i]=a[2*i-1].z=a[2*i].z;
            a[2*i].c=-1;
            a[2*i-1].c=1;
        }
        n*=2;
        sort(b+1,b+1+n);
        co=unique(b+1,b+1+n)-b-1;
        for(int i=1;i<=n;i++){
    
    
            a[i].yy=lower_bound(b+1,b+co+1,a[i].y)-b;
            a[i].zz=lower_bound(b+1,b+1+co,a[i].z)-b;
        }
        sort(a+1,a+1+n);
        build(1,1,co-1);
        for(int i=1;i<=n;i++){
    
    
            change(1,a[i].yy,a[i].zz-1,a[i].c);
            ans+=(a[i+1].x-a[i].x)*tree[1].dat;
        }
        printf("Test case #%lld\nTotal explored area: %.2f\n\n",++sum,ans);
    }
}