poj 1151 Atlantis

http://poj.org/problem?id=1151

这道题就是给你一些矩形的左上角和右下角的坐标，这些矩形可能有重叠，求这些矩形覆盖的面积。先把x坐标和y坐标分别离散化。然后再求面积。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #define maxn 510
 5 using namespace std;
 6 
 7 int n;
 8 double x2,y2,x1,y1;
 9 bool flag[maxn][maxn];
10 double X[maxn],Y[maxn];
11 struct node
12 {
13     double x1,y1,x2,y2;
14 } p[maxn];
15 
16 int bsearch(double *a,int l,int r,double target)
17 {
18     int low=l,high=r;
19     while(low<=high)
20     {
21         int mid=(low+high)>>1;
22         if(a[mid]==target)
23         {
24             return mid;
25         }
26         if(a[mid]>target)
27             high=mid-1;
28         else
29             low=mid+1;
30     }
31 }
32 
33 int main()
34 {
35     int case1=0;
36     while(scanf("%d",&n)!=EOF)
37     {
38         memset(flag,false,sizeof(flag));
39         memset(X,0,sizeof(X));
40         memset(Y,0,sizeof(Y));
41         case1++;
42         if(n==0) break;
43         int t1=0,t2=0;
44         for(int i=0; i<n; i++)
45         {
46             scanf("%lf%lf%lf%lf",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2);
47             X[t1++]=p[i].x1;X[t1++]=p[i].x2;
48             Y[t2++]=p[i].y1;Y[t2++]=p[i].y2;
49         }
50         sort(X,X+2*n);
51         sort(Y,Y+2*n);
52         for(int i=0; i<n; i++)
53         {
54             int xpos=bsearch(X,0,t1-1,p[i].x1);
55             int ypos=bsearch(Y,0,t2-1,p[i].y1);
56             int xpos1=bsearch(X,0,t1-1,p[i].x2);
57             int ypos1=bsearch(Y,0,t2-1,p[i].y2);
58             for(int i=xpos; i<xpos1; i++)
59             {
60                 for(int j=ypos; j<ypos1; j++)
61                 {
62                     flag[i][j]=true;
63                 }
64             }
65         }
66         double sum=0;
67         for(int i=0; i<t1; i++)
68         {
69             for(int j=0; j<t2; j++)
70             {
71                 if(flag[i][j])
72                     sum+=((X[i+1]-X[i])*(Y[j+1]-Y[j]));
73             }
74         }
75         printf("Test case #%d\n",case1);
76         printf("Total explored area: %.2lf\n",sum);
77         printf("\n");
78     }
79     return 0;
80 }
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posted @ 2014-02-20 21:20  null1019  阅读(168)  评论(0)  编辑  收藏  举报