POJ 1328 Radar Installation

题目：

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.

Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.

The input is terminated by a line containing pair of zeros

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

Examples

Input

Output

Case 1: 2

Case 2: 1

分析：

写完AB之后发现真的不知道该怎么写C了。。。。几乎没什么坑现在还能踩了，于是又加一个A
题意很好理解，理解完之后直接干，用的二维贪心，，，
然后

mle。。。。Wawawa
从网上找的题解，都和我不同的思路？？？？
wtf？？？？
我的思路没毛病啊？？
然后自己又仔细用数学证明了一遍，发现如果两个同样很远的点就会出现问题，我的贪心策略会使得灯塔++
emmmmmmmmm那就只能重打一遍了

错误代码：

 #include <stdio.h>

 #include <math.h>

 #include <string.h>

 #include <algorithm>

 #include <iostream>

 #include <string>

 #include <time.h>

 #include <queue>

 #include <string.h>

 #define sf scanf

 #define pf printf

 #define lf double

 #define ll long long

 #define p123 printf("123\n");

 #define pn printf("\n");

 #define pk printf(" ");

 #define p(n) printf("%d",n);

 #define pln(n) printf("%d\n",n);

 #define s(n) scanf("%d",&n);

 #define ss(n) scanf("%s",n);

 #define ps(n) printf("%s",n);

 #define sld(n) scanf("%lld",&n);

 #define pld(n) printf("%lld",n);

 #define slf(n) scanf("%lf",&n);

 #define plf(n) printf("%lf",n);

 #define sc(n) scanf("%c",&n);

 #define pc(n) printf("%c",n);

 #define gc getchar();

 #define re(n,a) memset(n,a,sizeof(n));

 #define len(a) strlen(a)

 #define LL long long

 #define eps 1e-6

 using namespace std;

 struct A {

     double x,y;

 } a[];

 int num = ;

 int n;

 double d;

 int count0 = ;

 bool cmp(A a, A b) {

     if(a.x < b.x)

         return true;

     if(fabs(a.x - b.x) <= eps && a.y > b.y)

         return true;

     return false;

 }

 int f(int x) {

     int xx = a[x].x + sqrt(d*d-a[x].y*a[x].y);

     num ++;

     for(int i = x+; i < n; i ++) { //p(i)

         if(sqrt( a[i].y*a[i].y + (a[i].x-xx)*(a[i].x-xx) ) > d) {

             f(i);

             return ;

         }

     }

     return ;

 }

 int main() {

     while(~scanf("%d%lf",&n,&d) && (n !=  || d != 0.0)) {

         num = ;

         count0 ++;

         int sta = ;

         for(int i = ; i < n; i ++) {

             slf(a[i].x) slf(a[i].y);

             if(a[i].y > d) {

                 sta = ;

             }

         }

         ps("Case ") p(count0) ps(": ");

         if(sta == ) {

             p(-) pn continue;

         }

         sort(a,a+n,cmp);

         f();

         p(num) pn

     }

     return ;

 }

AC代码：

 #include <stdio.h>

 #include <math.h>

 #include <string.h>

 #include <algorithm>

 #include <iostream>

 #include <string>

 #include <time.h>

 #include <queue>

 #include <string.h>

 #define sf scanf

 #define pf printf

 #define lf double

 #define ll long long

 #define p123 printf("123\n");

 #define pn printf("\n");

 #define pk printf(" ");

 #define p(n) printf("%d",n);

 #define pln(n) printf("%d\n",n);

 #define s(n) scanf("%d",&n);

 #define ss(n) scanf("%s",n);

 #define ps(n) printf("%s",n);

 #define sld(n) scanf("%lld",&n);

 #define pld(n) printf("%lld",n);

 #define slf(n) scanf("%lf",&n);

 #define plf(n) printf("%lf",n);

 #define sc(n) scanf("%c",&n);

 #define pc(n) printf("%c",n);

 #define gc getchar();

 #define re(n,a) memset(n,a,sizeof(n));

 #define len(a) strlen(a)

 #define LL long long

 #define eps 1e-6

 using namespace std;

 struct A {

     double l,r;

     int sta ;

 } a[];

 int n;

 double d;

 int count0 = ;

 int num = ;

 int count1 = ;

 bool cmp(A a, A b) {

     if(a.r!=b.r)

         return a.r<b.r;

     return a.l>b.l;

     return false;

 }

 int f(int x) {

     count1 ++;

     for(int i = x+; i < n; i++) {

         if(a[i].l > a[x].r) {

             f(i);

             return ;

         }

     }

     return ;

 }

 int main() {

     while(~scanf("%d%lf",&n,&d) && (n !=  || d != 0.0)) {

         num = ;

         count0 ++;

         count1 = ;

         int sta = ;

         double x,y;

         for(int i = ; i < n; i ++) {

             slf(x) slf(y);

             if(y > d) {

                 sta = ;

             }

             a[i].l = x-sqrt(d*d-y*y);

             a[i].r = x+ sqrt(d*d-y*y);

         }

         ps("Case ")

         p(count0)

         ps(": ");

         if(sta == ) {

             p(-) pn

             continue;

         }

         sort(a,a+n,cmp);

         //f(0);

         double end=-0x3f3f3f3f;//横坐标要很小....因为..你懂的

         for(int i=; i<n; i++) {

             if(end<a[i].l) {

                 end=a[i].r;

                 count1++;

             }

         }

         p(count1) pn

     }

     return ;

 }