poj 1328 贪心
贪心。以每个岛屿为圆心，在x轴上求出雷达可能出现的区间，重叠的区间只用一个雷达
1 #include <iostream>
2 #include <algorithm>
3 #include <cmath>
4 #include <stdio.h>
5 using namespace std;
6 typedef double ll;
7 const int maxx=1000;
8 struct PO
9 {
10 ll l,r;
11 };
12 PO point[maxx+5];
13
14 bool cmp(PO a,PO b)
15 {
16 if(a.l==b.l) return a.r<b.r;
17 return a.l<b.l;
18 }
19
20 int main()
21 {
22 int n,i,d,cnt,tt=0,flag;
23 ll a,b,ll,rr,temp1,temp2;
24 //freopen("in.txt","r",stdin);
25 while(scanf("%d%d",&n,&d)!=EOF)
26 {
27 if(n==0 && d==0) break;
28 temp1=(d*1.0)*(d*1.0);
29 flag=0;
30 for(i=0;i<n;i++)
31 {
32 scanf("%lf%lf",&a,&b);
33 temp2=b*b;
34 if(d-b<0)
35 flag=1;
36 if(flag) continue;
37 point[i].l=a-sqrt((temp1-temp2)*1.0);
38 point[i].r=a+sqrt((temp1-temp2)*1.0);
39 }
40 if(flag){printf("Case %d: -1\n",++tt);continue;}
41 sort(point,point+n,cmp);
42 ll=point[0].l;
43 rr=point[0].r;
44 cnt=1;
45 for(i=1;i<n;i++)
46 {
47 if(rr<point[i].l)
48 {
49 cnt++;
50 rr=point[i].r;
51 ll=point[i].l;
52 }
53 else if(rr>=point[i].l && rr<=point[i].r) ll=point[i].l;
54 else{ll=point[i].l;rr=point[i].r;}
55 }
56 printf("Case %d: %d\n",++tt,cnt);
57
58 }
59 return 0;
60 }
