POJ 1328 Radar Installation 贪心

把每个点转成区间，排序，首先考虑最左边的区间[l1,r1]，要覆盖住这个点则需要在这个区间里的某个位置x放一个雷达，显然尽量放在靠右端是最好的，因为可以同时被右边的其他区间覆盖，但是要被右边的区间[l2,r2]覆盖还需满足l2<=x<=r2，所以要x=min(x,r2)更新x的值。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("in.txt","r",stdin);
    //freopen("d:\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}

const int maxn = 10005;
const double eps = 10e-8;
typedef pair<double,double> pdd;
pdd da[maxn];

double cal(int x,int y,int R)
{
    if(abs(y)>R)return -1;
    return sqrt(R*R-y*y)+eps;
}
int main()
{
    int n,R;
    int ca=0;
    while(scanf("%d%d",&n,&R)!=EOF&&(n||R))
    {
        int sol=1;
        for(int i=0;i<n;i++)
        {
            int x,y; scanf("%d%d",&x,&y);
            double d=cal(x,y,R);
            if(d<=0){sol=0;continue;}
            da[i]=make_pair(x-d,x+d);
        }
        if(!sol) {printf("Case %d: %d
",++ca,-1);continue;}

        sort(da,da+n);
        double pre=-INT_MAX;
        int cnt=0;
        for(int i=0;i<n;)
        {
            while(i<n&&da[i].first<=pre){pre=min(pre,da[i].second);i++;}
            if(i==n)break;
            pre=da[i].second;
            cnt++;
            i++;
        }
        printf("Case %d: %d
",++ca,cnt);
    }
    return 0;
}