覆盖的面积 HDU - 1255 （求面积的交）

题目：给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积.

分析：在求面积并的基础上增加了一个成员len2，记录被覆盖至少两次的长度，num标记表示这一部分至少被全覆盖了几次（num = 1不一定代表这一段覆盖了只覆盖一次，还要看子节点）。同时考虑离散化。

代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 2010;
 4 struct edge
 5 {
 6     double x1, x2, h;
 7     int flag;
 8     edge(double a = 0, double b = 0, double c = 0, int d = 0) : x1(a), x2(b), h(c), flag(d){}
 9     bool operator< (const edge& b)const
10     {
11         return h < b.h;
12     }
13 }e[maxn];
14 
15 struct node
16 {
17     int l, r;
18     double len1, len2;
19     int num;
20     node(int a = 0, int b = 0, double c = 0, double d = 0, int e = 0) : l(a), r(b), len1(c), len2(d), num(e){}
21 }t[maxn << 2];
22 double x[maxn];
23 
24 void pushup(int tar)
25 {
26     if (t[tar].num)
27         t[tar].len1 = x[t[tar].r + 1] - x[t[tar].l];
28     else if (t[tar].l == t[tar].r)
29         t[tar].len1 = 0;
30     else t[tar].len1 = t[tar << 1].len1 + t[tar << 1 | 1].len1;
31     if (t[tar].num >= 2)
32         t[tar].len2 = x[t[tar].r + 1] - x[t[tar].l];
33     else if (t[tar].l == t[tar].r)
34         t[tar].len2 = 0;
35     else if (t[tar].num == 1)
36         t[tar].len2 = t[tar << 1].len1 + t[tar << 1 | 1].len1;
37     else t[tar].len2 = t[tar << 1].len2 + t[tar << 1 | 1].len2;
38 }
39 
40 void build(int l, int r, int tar)
41 {
42     t[tar].l = l, t[tar].r = r, t[tar].len1 = t[tar].len2 = 0, t[tar].num = 0;
43     if (l == r) return;
44     int mid = (l + r) >> 1;
45     build(l, mid, tar << 1), build(mid + 1, r, tar << 1 | 1);
46 }
47 
48 void update(int l, int r, int v, int tar)
49 {
50     //cout << tar << endl;
51     if (t[tar].l == l && t[tar].r == r)
52     {
53         t[tar].num += v;
54         pushup(tar);
55         return;
56     }
57     int mid = (t[tar].l + t[tar].r) >> 1;
58     if (r <= mid) update(l, r, v, tar << 1);
59     else if (l > mid) update(l, r, v, tar << 1 | 1);
60     else update(l, mid, v, tar << 1), update(mid + 1, r, v, tar << 1 | 1);
61     pushup(tar);
62 }
63 
64 int main()
65 {
66     int T; cin >> T;
67     int tot;
68     while (T--)
69     {
70         int n; cin >> n;
71 
72         tot = 0;
73         for (int i = 1; i <= n; i++)
74         {
75             double x1, y1, x2, y2;
76             scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
77             e[i * 2 - 1] = edge(x1, x2, y1, 1);
78             e[i * 2] = edge(x1, x2, y2, -1);
79             x[i * 2 - 1] = x1, x[i * 2] = x2;
80         }
81         sort(e + 1, e + 1 + 2 * n);
82         sort(x + 1, x + 1 + 2 * n);
83         tot = unique(x + 1, x + 1 + 2 * n) - x - 1;
84         //cout << tot << endl;
85         build(1, tot - 1, 1);
86         double res = 0;
87         for (int i = 1; i < 2 * n; i++)
88         {
89             int x1 = lower_bound(x + 1, x + 1 + tot, e[i].x1) - x;
90             int x2 = lower_bound(x + 1, x + 1 + tot, e[i].x2) - x;
91             //cout << x1 << " " << x2 << endl;
92             update(x1, x2 - 1, e[i].flag, 1);
93             res += (e[i + 1].h - e[i].h) * t[1].len2;
94         }
95         printf("%.2f\n", res);
96     }
97 }