#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>#define lson rt<<1,l,mid
#define rson rt<<1|1,mid + 1,r
using namespace std;
const int maxn = 4004;struct Seg {double l, r, x;int cover;Seg(double l, double r, double x, int cover) :l(l), r(r), x(x), cover(cover) {}//排序大小按照x轴的坐标大小bool operator < (const Seg &tmp) const {return x < tmp.x;}
};int n, cnt[maxn << 2];
double len[maxn << 2];
vector<double> dy;
vector<Seg> seg;int GetID(double Val) {//二分查找点的坐标，离散化return lower_bound(dy.begin(), dy.end(), Val) - dy.begin();
}void pushup(int rt, int l, int r) {int lc = rt << 1, rc = rt << 1 | 1;//如果当前线段存在，更新坐标//这里r+1是因为长度为x的线段有x+1个点，所以最后一个点是r+1if (cnt[rt]) len[rt] = dy[r + 1] - dy[l];else if (r == l) len[rt] = 0; else len[rt] = len[lc] + len[rc];
}void update(int rt, int l, int r, int ql, int qr, int Val) {int lc = rt << 1, rc = rt << 1 | 1, mid = (l + r) >> 1;if (ql <= l && qr >= r) {//累加当前线段的值cnt[rt] += Val;pushup(rt, l, r);}else {//标记不需要下放，因为当前线段的增减和子线段没有关系//而且每条线段必定增加一次，减少一次if (ql <= mid) update(lson, ql, qr, Val);if (qr > mid) update(rson, ql, qr, Val);pushup(rt, l, r);}
}int main() {int kase = 1;while (scanf("%d", &n),n != 0) {//初始化线段树dy.clear(); seg.clear();memset(cnt, 0, sizeof(cnt));memset(len, 0, sizeof(len));//读入数据for (int i = 1; i <= n; i++) {double x1, x2, y1, y2; scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);//将每个矩形抽象成两条线段seg.push_back(Seg(y1, y2, x1, 1));seg.push_back(Seg(y1, y2, x2, -1));//存储每个坐标的值dy.push_back(y1); dy.push_back(y2);}//将浮点坐标离散化sort(dy.begin(), dy.end());dy.erase(unique(dy.begin(), dy.end()), dy.end());//将线段排序sort(seg.begin(), seg.end());int msize = seg.size(), k = dy.size();double ret = 0;for (int i = 0; i < msize - 1; i++) {//这里的qr为什么要减一？ 是因为我线段树每个节点存的是l,r之间的线段，而不是点集，把每个线段当成点来看的话，//每个[l,r]的点构成的线段编号应该是[l,r-1]int ql = GetID(seg[i].l), qr = GetID(seg[i].r) - 1;//更新值update(1, 0, k - 1, ql, qr, seg[i].cover);//累加当前扫描线扫过的面积ret += len[1] * (seg[i + 1].x - seg[i].x);}printf("Test case #%d\nTotal explored area: %.2lf\n\n", kase++, ret);}return 0;
}