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
題目的意思是給定n個(gè)矩形的2n個(gè)坐標(biāo)，求矩形的覆蓋面積。如果開(kāi)一個(gè)大的bool數(shù)組，將覆蓋過(guò)的部分更新為true，再?gòu)念^到尾掃描一遍，在坐標(biāo)范圍比較小的情況下，可以求解。但是如果坐標(biāo)x,y的取值范圍很大，比如[-10^8,10^8]，用上面這個(gè)方法就不能求解了；而且坐標(biāo)還有可能是實(shí)數(shù)，上面的方法就更加不可行了，需要尋找一種新的解法，就是下面要說(shuō)到的“離散化”。
注意到要表示一個(gè)矩形，只需要知道其2個(gè)頂點(diǎn)的坐標(biāo)就可以了(最左下，最右上)。可以用2個(gè)數(shù)組x[0...2n-1]，y[0...2n-1]記錄下矩形Ri的2個(gè)坐標(biāo)(x1,y1)，(x2,y2)，然后將數(shù)組x[0...xn-1]，y[0...2n-1]排序，為下一步的掃描線作準(zhǔn)備，這就是離散化的思想。這題還可以用線段樹(shù)做進(jìn)一步優(yōu)化，但是這里只介紹離散化的思想。
看下面這個(gè)例子：有2個(gè)矩形(1,1)，(3,3)和(2,2)，(4,4)。如圖：
圖中虛線表示掃描線，下一步工作只需要將這2個(gè)矩形覆蓋過(guò)的部分的bool數(shù)組的對(duì)應(yīng)位置更新為true，接下去用掃描線從左到右，從上到下掃描一遍，就可以求出矩形覆蓋的總面積。
這個(gè)圖對(duì)應(yīng)的bool數(shù)組的值如下：
1 1 0 1 2 3
1 1 1 <----> 4 5 6
0 1 1 7 8 9
1 #include <iostream>
2 #include <cmath>
3 using namespace std;
4
5 const int N = 101;
6 const double eps = 1e-6;
7 double ans,x[2*N],y[2*N];
8 double pos[N][4];
9 bool hash[2*N][2*N];
10
11 int cmp(const void *a,const void *b){
12 double *aa = (double *)a;
13 double *bb = (double *)b;
14 if(fabs(*aa-*bb)<=eps) return 0;
15 else if(*aa-*bb>0) return 1;
16 else return -1;
17 }
18 int main(){
19 int i,j,k,n,x1,x2,y1,y2,ca=1;
20 while(scanf("%d",&n),n){
21 for(ans=i=k=0;i<n;i++,k+=2){
22 scanf("%lf %lf %lf %lf",&pos[i][0],&pos[i][1],&pos[i][2],&pos[i][3]);
23 x[k]=pos[i][0],y[k]=pos[i][1],x[k+1]=pos[i][2],y[k+1]=pos[i][3];
24 }
25 memset(hash,false,sizeof(hash));
26 qsort(x,2*n,sizeof(x[0]),cmp);
27 qsort(y,2*n,sizeof(y[0]),cmp);
28 for(i=0;i<n;i++){
29 for(k=0;fabs(x[k]-pos[i][0])>eps;k++); x1=k;
30 for(k=0;fabs(y[k]-pos[i][1])>eps;k++); y1=k;
31 for(k=0;fabs(x[k]-pos[i][2])>eps;k++); x2=k;
32 for(k=0;fabs(y[k]-pos[i][3])>eps;k++); y2=k;
33 for(j=x1;j<x2;j++) for(k=y1;k<y2;k++)
34 hash[j][k]=true;
35 }
36 for(i=0;i<2*n-1;i++)
37 for(j=0;j<2*n-1;j++)
38 ans+=hash[i][j]*(x[i+1]-x[i])*(y[j+1]-y[j]);
39 printf("Test case #%d\n",ca++);
40 printf("Total explored area: %.2lf\n",ans);
41 printf("\n");
42 }
43 return 0;
44 }