zoj1360/poj1328 Radar Installation（贪心）

对每个岛屿，能覆盖它的雷达位于线段[x-sqrt(d*d-y*y),x+sqrt(d*d+y*y)],那么把每个岛屿对应的线段求出来后，其实就转化成了经典的贪心法案例：区间选点问题。数轴上有n个闭区间[ai,bi],取尽量少的点，使得每个区间内都至少有一个点。选法是：把区间按右端点从小到大排序（右端点相同时按左端点从大到小），然后每个没选的区间选最右边的一个点即可。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<list>
#include<deque>
#include<vector>
#include<algorithm>
#include<stack>
#include<queue>
#include<cctype>
#include<sstream>
using namespace std;
#define pii pair<int,int>
#define LL long long int
const double eps=1e-10;
const int INF=1000000000;
const int maxn=1000+10;

int n,cas=1;
double d,x,y;
struct Line
{
    double l,r;
    bool used;
    Line(double ll=0,double rr=0,bool uu=false):l(ll),r(rr),used(uu) {}
    Line(const Line& C)
    {
        l=C.l;
        r=C.r;
        used=C.used;
    }
    bool operator < (const Line& B) const
    {
        if(r!=B.r)
            return r<B.r;
        else
            return l>B.l;
    }
} a[maxn];

int main()
{
    //freopen("in2.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%d%lf",&n,&d)==2)
    {
        if(n==0&&d==0) break;
        printf("Case %d: ",cas++);
        int ans=0;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf",&x,&y);
            if(y>d)
            {
                ans=-1;
                break;
            }
            a[i].l=x-sqrt(d*d-y*y);
            a[i].r=x+sqrt(d*d-y*y);
            //cout<<a[i].l<<‘ ‘<<a[i].r<<endl;
            a[i].used=false;
        }
        if(ans==-1)
        {
            printf("-1\n");
        }
        else
        {
            sort(a,a+n);
            for(int i=0; i<n; i++)
            {
                //cout<<a[i].l<<‘ ‘<<a[i].r<<endl;
                if(a[i].used==false)
                {
                    a[i].used=true;
                    ans++;
                    for(int j=i+1; j<n; j++)
                    {
                        if(a[j].l<=a[i].r)
                        {
                            a[j].used=true;
                        }
                        else break;
                    }
                }
            }
            printf("%d\n",ans);
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}