HDU - 1542【離散化+線段樹+掃描線】

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11997    Accepted Submission(s): 5056

Problem Description

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

 

 

Input

The input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.

The input file is terminated by a line containing a single 0. Don’t process it.

 

 

Output

For each test case, your program should output one section. The first line of each section must be 「Test case #k」, where k is the number of the test case (starting with 1). The second one must be 「Total explored area: a」, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.

Output a blank line after each test case.

 

 

Sample Input

2 10 10 20 20 15 15 25 25.5 0

 

 

Sample Output

Test case #1 Total explored area: 180.00

 

題意：

給你每一個矩形的左下和右上的點，讓你求面積並。

題解：

裸地掃描線+線段樹維護區間和。我是從左往右掃，所以離散化的是y座標。注意區間更新的時候，由於點數爲cnt的點實際上維護的是長度爲cnt-1的區間，

所以更新的時候是 pos1+1, pos2;

當時在學校認真思考了一早上，一直懶得補，今天寫了以後發現離散化寫的很煩，本身太挫了，後來發現別人的很清晰，就拿來用了。

代碼：

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <map>
 6 #include <vector>
 7 #include <queue>
 8 #include <list>
 9 #include <cstdio>
10 #define rep(i,a,b) for(int (i) = (a);(i) <= (b);++ (i))
11 #define per(i,a,b) for(int (i) = (a);(i) >= (b);-- (i))
12 #define mem(a,b) memset((a),(b),sizeof((a)))
13 #define FIN freopen("in.txt","r",stdin)
14 #define FOUT freopen("out.txt","w",stdout)
15 #define IO ios_base::sync_with_stdio(0),cin.tie(0)
16 #define mid ((l+r)>>1)
17 #define ls (id<<1)
18 #define rs ((id<<1)|1)
19 #define INF 0x3f3f3f3f
20 typedef long long ll;
21 const ll mod = 1e8+7;
22 const ll eps = 1e-12;
23 using namespace std;
24 const int N = 210;
25 
26 int n;
27 double xp1, yp1, xp2, yp2, y[N*4];
28 struct Line{
29     double x, y1, y2;
30     int f;
31     bool operator < (const Line &r) const {
32         return x < r.x;
33     }
34 }hashy[N*4];
35 struct Node{
36     double sum;
37     int lazy;
38 }node[N*4];
39 
40 void pushUp(int id, int l, int r){
41     if(node[id].lazy > 0)    node[id].sum = y[r-1]-y[l-2];
42     else if(l == r) node[id].sum = 0;
43     else
44         node[id].sum = node[ls].sum+node[rs].sum;
45 }
46 void build(int id, int l, int r){
47     node[id].lazy = 0;
48     if(l == r){
49         node[id].sum = 0;
50         return ;
51     }
52     build(ls, l, mid);
53     build(rs, mid+1, r);
54 }
55 void update(int id, int l, int r, int ql, int qr, int p){
56     if(ql == l && qr == r){
57         node[id].lazy += p;
58         pushUp(id, l, r);
59         return ;
60     }
61     if(qr <= mid)   update(ls, l, mid, ql, qr, p);
62     else if(ql > mid)
63         update(rs, mid+1, r, ql, qr, p);
64     else{
65         update(ls, l, mid, ql, mid, p);
66         update(rs, mid+1, r, mid+1, qr, p);
67     }
68     pushUp(id, l, r);
69 }
70 int main()
71 {
72     //FIN;
73     int w_w = 0;
74     while(cin >> n, n != 0){
75         int len = 0;
76         rep(i, 1, n){
77             cin >> xp1 >> yp1 >> xp2 >> yp2;
78             hashy[len].x = xp1; hashy[len].y1 = yp1;
79             hashy[len].y2 = yp2; hashy[len].f = 1; y[len++] = yp1;
80             hashy[len].x = xp2; hashy[len].y1 = yp1;
81             hashy[len].y2 = yp2; hashy[len].f = -1; y[len++] = yp2;
82         }
83         sort(hashy, hashy+len);
84         sort(y, y+len);
85 
86         build(1, 1, n*2);
87         double ans = 0;
88         rep(i, 0, len-1){
89             int pos1 = lower_bound(y, y+len, hashy[i].y1)-y+1;
90             int pos2 = lower_bound(y, y+len, hashy[i].y2)-y+1;
91             if(i)   ans += node[1].sum*(hashy[i].x-hashy[i-1].x);
92             update(1, 1, n*2, pos1+1, pos2, hashy[i].f);
93             //cout << "sum: " << node[1].sum << endl;
94         }
95         cout << "Test case #" << ++w_w << endl;
96         printf("Total explored area: %.2lf\n\n", ans);
97     }
98     return 0;
99 }

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