HDU 1542 线段树+扫描线+离散化

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.

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

10 10 20 20

15 15 25 25.5

Sample Output

Test case #1

Total explored area: 180.00

Source

Mid-Central European Regional Contest 2000

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 }