Atlantis
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8327 Accepted Submission(s): 3627
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.
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.
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
Source
题目意思：
给出n个矩形，求总面积（覆盖的只算一次）
思路：
扫描线模板题
代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <vector> 6 #include <queue> 7 #include <cmath> 8 #include <set> 9 using namespace std; 10 11 #define N 105 12 #define ll root<<1 13 #define rr root<<1|1 14 #define mid (a[root].l+a[root].r)/2 15 16 int max(int x,int y){return x>y?x:y;} 17 int min(int x,int y){return x<y?x:y;} 18 int abs(int x,int y){return x<0?-x:x;} 19 20 struct node{ 21 int l, r; 22 double sum; 23 int val; 24 }a[N*8]; 25 26 struct Line{ 27 double x1, x2, y; 28 int val; 29 Line(){} 30 Line(double a,double b,double c,int d){ 31 x1=a; 32 x2=b; 33 y=c; 34 val=d; 35 } 36 }line[N*2]; 37 38 bool cmp(Line a,Line b){ 39 return a.y<b.y; 40 } 41 int n, m; 42 double xx[N*2]; 43 44 int b_s(double key){ 45 int l=1, r=m; 46 while(l<=r){ 47 int mm=(l+r)/2; 48 if(xx[mm]==key) return mm; 49 else if(xx[mm]>key) r=mm-1; 50 else if(xx[mm]<key) l=mm+1; 51 } 52 } 53 54 void build(int l,int r,int root){ 55 a[root].l=l; 56 a[root].r=r; 57 a[root].sum=a[root].val=0; 58 if(l==r) return; 59 build(l,mid,ll); 60 build(mid+1,r,rr); 61 } 62 63 void up(int root){ 64 if(a[root].val) a[root].sum=xx[a[root].r+1]-xx[a[root].l]; 65 else if(a[root].l==a[root].r) a[root].sum=0; 66 else a[root].sum=a[ll].sum+a[rr].sum; 67 68 } 69 70 void update(int l,int r,int val,int root){ 71 if(a[root].l==l&&a[root].r==r){ 72 a[root].val+=val; 73 up(root); 74 return; 75 } 76 if(l>=a[rr].l) update(l,r,val,rr); 77 else if(r<=a[ll].r) update(l,r,val,ll); 78 else{ 79 update(l,mid,val,ll); 80 update(mid+1,r,val,rr); 81 } 82 up(root); 83 } 84 85 main() 86 { 87 int i, j, k; 88 double x1, y1, x2, y2; 89 int kase=1; 90 while(scanf("%d",&n)==1&&n){ 91 m=1;k=0; 92 for(i=0;i<n;i++){ 93 scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2); 94 line[k++]=Line(x1,x2,y1,1); 95 line[k++]=Line(x1,x2,y2,-1); 96 xx[m++]=x1;xx[m++]=x2; 97 } 98 sort(xx+1,xx+m); 99 m=unique(xx+1,xx+m)-xx-1; 100 sort(line,line+k,cmp); 101 build(1,m,1); 102 double ans=0.0; 103 for(i=0;i<k-1;i++){ 104 update(b_s(line[i].x1),b_s(line[i].x2)-1,line[i].val,1); 105 ans+=a[1].sum*(line[i+1].y-line[i].y); 106 } 107 printf("Test case #%d ",kase++); 108 printf("Total explored area: %.2f ",ans); 109 } 110 }