poj1328

一、题意：有n个小岛，坐标为(x,y)。以x轴为海岸线，在海岸线上布置雷达，雷达能覆盖半径为d的圆形区域。求最少用多少个雷达能覆盖所有的小岛

 

二、思路：以小岛为圆心，d为半径作圆，其与x轴会有两个交点。这两个交点间的线段，就是满足这题小岛要求的雷达坐标。然后将从这个线段从左到右排序，有交集的线段就表示这两个小岛可以共用一个雷达，从而转换成一个区间贪心的问题。

 

三、代码：

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#include"iostream" #include"stdio.h" #include"algorithm" #include"cmath" using namespace std; int n,d; struct Node {     double x,y; }; struct InterSection {     double x1,x2; }; Node cor[1005]; InterSection segment[1005];   bool Cmp(const InterSection a,const InterSection b) {     if(a.x2!=b.x2) return a.x2<b.x2;     else return a.x1>b.x1; }   void CalIntersection(int i) {     segment[i].x1=sqrt(d*d-cor[i].y*cor[i].y)+cor[i].x;     segment[i].x2=cor[i].x-sqrt(d*d-cor[i].y*cor[i].y); }   double Min(double a,double b) {     return a<b?a:b; }   int Solve() {     double left=segment[0].x2,right=segment[0].x1;     int ans=1,i=1;     while(i<n)     {         while(i<n&&segment[i].x2<=right)         {             left=segment[i].x2;             right=Min(right,segment[i].x1);             i++;         }         if(i<n)         {             ans++;             left=segment[i].x2;             right=segment[i].x1;         }     }     return ans; } int main() {     int cnt=0;     while(cin>>n>>d,n||d)     {         int ans=0;         for(int i=0;i<n;i++)         {             cin>>cor[i].x>>cor[i].y;             if(cor[i].y>d) ans=-1;             CalIntersection(i);         }         if(ans!=-1){             sort(segment,segment+n,Cmp);             ans=Solve();         }         cout<<"Case "<<++cnt<<": "<<ans<<endl;     }     return 0; }