Atlantis（线段树+扫描线）

Atlantis

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. 

InputThe 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.OutputFor 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 
题意：给出矩形的左下和右上两个端点的横纵坐标，求出所有矩形的面积，覆盖部分算一次。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 const int maxn=1100;
 7 struct node{
 8     double l,r,val;
 9     int f,d;
10 }tree[maxn<<2];
11 struct node2{
12     double l,r,h;
13     int f;
14 }e[maxn<<2];
15 double a[maxn<<2];
16 int n;
17 bool cmp(node2 a,node2 b)
18 {
19     return a.h<b.h;
20 }
21 void pushup(int cur)
22 {
23     if(tree[cur].f>0)//就比如说我要更新10-20，现在处于10-15的区间即cur=2 
24     {
25         tree[cur].val=(tree[cur].r-tree[cur].l); 
26     }
27     else if(tree[cur].d==1)
28     {
29         tree[cur].val=0;
30     }
31     else
32     {
33         tree[cur].val=tree[cur*2].val+tree[cur*2+1].val;//例如当cur=1,tree[cur].val就等于其左子树10-15的val值加上右子树保存的15-20的val值 
34     }
35 }
36 void build(int l,int r,int cur)//第一次调用l=1 r=4相当于用区间1-4表示横坐标10-25的区间 
37 {
38     tree[cur].l=a[l];
39     tree[cur].r=a[r];
40     tree[cur].f=0;
41     tree[cur].val=0;
42     if(l+1==r)//离散化后的区间最小值，比如2-3就是一个线段的最小值，3-3就只能表示一个点 
43     {
44         tree[cur].d=1;//相当于已经到了叶子节点 
45         return ;//详见下图（1） 
46     }
47     tree[cur].d=0;
48     int mid=(l+r)/2;
49     build(l,mid,cur*2);//因为是区间1-3 应该分为1-2和2-3 
50     build(mid,r,cur*2+1);//所以mid不用加一 
51 }
52 void update(double pl,double pr,int f,int cur)
53 {
54     if(pl<=tree[cur].l&&tree[cur].r<=pr)
55     {
56         tree[cur].f+=f; 
57         pushup(cur);
58         return ; 
59     }
60     if(pl<tree[cur*2].r) update(pl,pr,f,cur*2);
61     if(pr>tree[cur*2+1].l) update(pl,pr,f,cur*2+1);
62     pushup(cur);
63 }
64 int main()
65 {
66     int T=1;
67     while(~scanf("%d",&n)&&n)
68     {
69         double x1,x2,y1,y2;
70         for(int i=1;i<=n;i++)
71         {
72             scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
73             e[i*2-1].l=x1;e[i*2-1].r=x2;//记录每一条线段的左右端点 
74             e[i*2-1].h=y1;e[i*2-1].f=1;//f=1表示进入该矩形 
75             e[i*2].l=x1;e[i*2].r=x2;
76             e[i*2].h=y2;e[i*2].f=-1;//f=-1表示出该矩形 
77             a[2*i-1]=x1;
78             a[2*i]=x2;
79         } 
80         sort(a+1,a+1+n*2);//将横坐标从小到大排序，为将其离散化做准备 
81         int k=1;
82         for(int i=2;i<=n*2;i++)
83         {
84             if(a[i]!=a[i-1])
85                 a[++k]=a[i];
86         }//去重并离散化 例如题目样例对应a[1]=10 a[2]=15  a[3]=20 a[4]=25
87         build(1,k,1);
88         sort(e+1,e+1+2*n,cmp);//将矩形按高度从小到大排序，因为扫描线是从下往上扫 
89         double ans=0;
90         for(int i=1;i<=n*2;i++)
91         {
92             ans+=(e[i].h-e[i-1].h)*tree[1].val;
93             update(e[i].l,e[i].r,e[i].f,1);
94         }
95         printf("Test case #%d\nTotal explored area: %.2lf\n\n",T++,ans);
96     } 
97 } 

 

转载于:https://www.cnblogs.com/1013star/p/9445566.html