1.题目链接。其实是一个比较经典的问题，自己对着代码手动的画一下就明白了。主要是弄清楚线段树维护的到底是个什么东西，在扫描线向上扫描的过程中，线段树维护了当前x轴上有贡献的线段长度。具体的思路是首先离散化每个点，然后在这些点上打标记，记录每个点的贡献，算出当前线段的长度即可。

#include<bits/stdc++.h>
#include <algorithm>
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;
const int N = 205, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n;
struct Seg 
{
	double l, r, h;
	int d;
	Seg() {}
	Seg(double l, double r, double h, int d) : l(l), r(r), h(h), d(d) {}
	bool operator< (const Seg& rhs) const { return h < rhs.h; }
} a[N];

int cnt[N << 2];
double sum[N << 2], all[N];


void push_up(int l, int r, int rt) 
{
	if (cnt[rt]) sum[rt] = all[r + 1] - all[l];
	else if (l == r) sum[rt] = 0;
	else sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void update(int L, int R, int v, int l, int r, int rt) 
{
	if (L <= l && r <= R) {
		cnt[rt] += v;
		push_up(l, r, rt);
		return;
	}
	int m = l + r >> 1;
	if (L <= m) update(L, R, v, lson);
	if (R > m) update(L, R, v, rson);
	push_up(l, r, rt);
}

int main() 
{
	int kase = 0;
	while (scanf("%d", &n) == 1 && n) 
	{
		for (int i = 1; i <= n; ++i) 
		{
			double x1, y1, x2, y2;
			scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
			a[i] = Seg(x1, x2, y1, 1);
			a[i + n] = Seg(x1, x2, y2, -1);
			all[i] = x1; all[i + n] = x2;
		}
		n <<= 1;
		sort(a + 1, a + 1 + n);
		sort(all + 1, all + 1 + n);
		int m = unique(all + 1, all + 1 + n) - all - 1;

		memset(cnt, 0, sizeof cnt);
		memset(sum, 0, sizeof sum);
		double ans = 0;
		for (int i = 1; i < n; ++i) {
			int l = lower_bound(all + 1, all + 1 + m, a[i].l) - all;
			int r = lower_bound(all + 1, all + 1 + m, a[i].r) - all;
			if (l < r) update(l, r - 1, a[i].d, 1, m, 1);
			ans += sum[1] * (a[i + 1].h - a[i].h);
		}
		printf("Test case #%d\nTotal explored area: %.2f\n\n", ++kase, ans);
	}
	return 0;
}