区间贪心·POJ 1328·Radar Installation

题目大意：

坐标系中有n个小岛，在x轴上设置雷达，雷达范围是d，要求用最少雷达能覆盖所有小岛；

解题思路：

如果直接用x轴上点到岛的距离进行计算，不仅麻烦而且很容易打乱思路，毕竟圆形范围不是很好处理。

我们可以把它转化为x轴上的区间问题，既然岛到雷达距离不能超过d，那么在x轴上必定对应一个范围使得雷达必须在区间内才可以覆盖到岛。

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <fstream>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;
#define   maxn          1010
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   ms(x,y)      memset(x,y,sizeof(x))
#define   rep(i,n)      for(int i=0;i<(n);i++)
#define   repf(i,a,b)   for(int i=(a);i<=(b);i++)
#define   pii           pair<int,int>
//#define   mp            make_pair
#define   FI            first
#define   SE            second
#define   IT            iterator
#define   PB            push_back
#define   Times         10
typedef   long long     ll;
typedef   unsigned long long ull;
typedef   long double   ld;
typedef   pair<int ,int > P;

const double eps = 1e-10;
const double  pi = acos(-1.0);
const  ll    mod = 1e9+7;
const  int   inf = 0x3f3f3f3f;
const  ll    INF = (ll)1e18+300;
const int   maxd = 1000 + 10;

int n;
double d;
struct  node{
    double l, r;
    node(double xx = 0, double yy = 0): l(xx), r(yy){}
};
node ac[maxd];

bool cmp(node a, node b) {
    if(a.l != b.l) {
        return a.l < b.l;
    }
    else return a.r > b.r;
}
int main() {
    int kase = 0;
    while(cin >> n >> d && (n +d)) {
        kase ++;
        double max_d = -1;
        int flag = 1;
        for (int i = 0; i < n; i++){
            double xx, yy;
            cin >> xx >> yy;
            if(flag == 0) {
                continue;
            }
            if(yy > d) {
                flag = 0;
                continue;
            }
            ac[i].l = xx - sqrt(d*d - yy*yy);
            ac[i].r = xx + sqrt(d*d - yy*yy);
        }
        if(!flag) {
            cout << "Case " << kase << ": ";
            cout << "-1" << endl;
            continue;
        }
        sort(ac, ac + n, cmp);
        double max_r = ac[0].r;
        int f = 0;
        int ans = 1;
        for (int i = 1; i < n; i++) {
            if(ac[i].l <= max_r) {
                if(ac[i].r < max_r) {
                    max_r = ac[i].r;
                }
            }
            else{
                max_r = ac[i].r;
                ans ++;
            }
        }
        cout<<"Case "<<kase<<": "<<ans<<endl;
    }
}