大意: 求矩形面积并.
枚举$x$坐标, 线段树维护$[y_1,y_2]$内的边是否被覆盖, 线段树维护边时需要将每条边挂在左端点上.
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head
const int N = 1e3+10;
int n, cas, cnt, tot;
struct _ {double l,r,h;int v;} e[N];
double b[N], sum[N<<2];
int tag[N<<2];
void pu(int o, int l, int r) {
if (tag[o]) sum[o] = b[r+1]-b[l];
else sum[o] = sum[lc]+sum[rc];
}
void update(int o, int l, int r, int ql, int qr, int v) {
if (ql<=l&&r<=qr) return tag[o]+=v,pu(o,l,r);
if (mid>=ql) update(ls,ql,qr,v);
if (mid<qr) update(rs,ql,qr,v);
pu(o,l,r);
}
void work() {
cnt=tot=0;
memset(sum,0,sizeof sum);
memset(tag,0,sizeof tag);
REP(i,1,n) {
double x1, y1, x2, y2;
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
e[++cnt]={y1,y2,x1,1};
e[++cnt]={y1,y2,x2,-1};
b[++tot]=y1,b[++tot]=y2;
}
sort(b+1,b+1+tot),tot=unique(b+1,b+1+tot)-b-1;
sort(e+1,e+1+cnt,[](_ a,_ b){return a.h<b.h;});
double ans = 0;
REP(i,1,cnt-1) {
int l=lower_bound(b+1,b+1+tot,e[i].l)-b;
int r=lower_bound(b+1,b+1+tot,e[i].r)-b-1;
update(1,1,tot,l,r,e[i].v);
ans += sum[1]*(e[i+1].h-e[i].h);
}
printf("Test case #%d\nTotal explored area: %.2lf\n\n",++cas,ans);
}
int main() {
for (; scanf("%d", &n), n; ) work();
}