POJ 1328 Radar Installation

https://vjudge.net/problem/POJ-1328

題目：

    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

Examples

Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Output

Case 1: 2
Case 2: 1

分析：

    寫完AB之後發現真的不知道該怎麼寫C了。。。。幾乎沒什麼坑現在還能踩了，於是又加一個A
    題意很好理解，理解完之後直接幹，用的二維貪心，，，
    然後

mle。。。。Wawawa
從網上找的題解，都和我不同的思路？？？？
wtf？？？？
我的思路沒毛病啊？？
然後自己又仔細用數學證明了一遍，發現如果兩個同樣很遠的點就會出現問題，我的貪心策略會使得燈塔++
emmmmmmmmm那就只能重打一遍了

錯誤程式碼：

 1 #include <stdio.h>
 2 #include <math.h> 
 3 #include <string.h> 
 4 #include <algorithm> 
 5 #include <iostream>
 6 #include <string> 
 7 #include <time.h> 
 8 #include <queue> 
 9 #include <string.h>
10 #define sf scanf 
11 #define pf printf 
12 #define lf double 
13 #define ll long long
14 #define p123 printf("123\n"); 
15 #define pn printf("\n"); 
16 #define pk printf(" ");
17 #define p(n) printf("%d",n); 
18 #define pln(n) printf("%d\n",n); 
19 #define s(n) scanf("%d",&n);
20 #define ss(n) scanf("%s",n); 
21 #define ps(n) printf("%s",n); 
22 #define sld(n) scanf("%lld",&n);
23 #define pld(n) printf("%lld",n); 
24 #define slf(n) scanf("%lf",&n);
25 #define plf(n) printf("%lf",n); 
26 #define sc(n) scanf("%c",&n);
27 #define pc(n) printf("%c",n); 
28 #define gc getchar();
29 #define re(n,a) memset(n,a,sizeof(n));
30 #define len(a) strlen(a) 
31 #define LL long long 
32 #define eps 1e-6
33 using namespace std;
34 struct A {
35     double x,y;
36 } a[100000];
37 int num = 0;
38 int n;
39 double d;
40 int count0 = 0;
41 bool cmp(A a, A b) {
42     if(a.x < b.x)
43         return true;
44     if(fabs(a.x - b.x) <= eps && a.y > b.y)
45         return true;
46     return false;
47 }
48 int f(int x) {
49     int xx = a[x].x + sqrt(d*d-a[x].y*a[x].y);
50     num ++;
51     for(int i = x+1; i < n; i ++) { //p(i)
52         if(sqrt( a[i].y*a[i].y + (a[i].x-xx)*(a[i].x-xx) ) > d) {
53             f(i);
54             return 0;
55         }
56     }
57     return 0;
58 }
59 int main() {
60     while(~scanf("%d%lf",&n,&d) && (n != 0 || d != 0.0)) {
61         num = 0;
62         count0 ++;
63         int sta = 0;
64         for(int i = 0; i < n; i ++) {
65             slf(a[i].x) slf(a[i].y);
66             if(a[i].y > d) {
67                 sta = 1;
68             }
69         }
70         ps("Case ") p(count0) ps(": ");
71         if(sta == 1) {
72             p(-1) pn continue;
73         }
74         sort(a,a+n,cmp);
75         f(0);
76         p(num) pn
77     }
78     return 0;
79 }

AC程式碼：

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <iostream>
 6 #include <string>
 7 #include <time.h>
 8 #include <queue>
 9 #include <string.h>
10 #define sf scanf
11 #define pf printf
12 #define lf double
13 #define ll long long
14 #define p123 printf("123\n");
15 #define pn printf("\n");
16 #define pk printf(" ");
17 #define p(n) printf("%d",n);
18 #define pln(n) printf("%d\n",n);
19 #define s(n) scanf("%d",&n);
20 #define ss(n) scanf("%s",n);
21 #define ps(n) printf("%s",n);
22 #define sld(n) scanf("%lld",&n);
23 #define pld(n) printf("%lld",n);
24 #define slf(n) scanf("%lf",&n);
25 #define plf(n) printf("%lf",n);
26 #define sc(n) scanf("%c",&n);
27 #define pc(n) printf("%c",n);
28 #define gc getchar();
29 #define re(n,a) memset(n,a,sizeof(n));
30 #define len(a) strlen(a)
31 #define LL long long
32 #define eps 1e-6
33 using namespace std;
34 struct A {
35     double l,r;
36     int sta ;
37 } a[10000];
38 int n;
39 double d;
40 int count0 = 0;
41 int num = 0;
42 int count1 = 0;
43 bool cmp(A a, A b) {
44     if(a.r!=b.r)
45         return a.r<b.r;
46     return a.l>b.l;
47     return false;
48 }
49 
50 int f(int x) {
51     count1 ++;
52     for(int i = x+1; i < n; i++) {
53         if(a[i].l > a[x].r) {
54             f(i);
55             return 0;
56         }
57     }
58     return 0;
59 }
60 
61 int main() {
62     while(~scanf("%d%lf",&n,&d) && (n != 0 || d != 0.0)) {
63         num = 0;
64         count0 ++;
65         count1 = 0;
66         int sta = 0;
67         double x,y;
68         for(int i = 0; i < n; i ++) {
69             slf(x) slf(y);
70             if(y > d) {
71                 sta = 1;
72             }
73             a[i].l = x-sqrt(d*d-y*y);
74             a[i].r = x+ sqrt(d*d-y*y);
75         }
76         ps("Case ")
77         p(count0)
78         ps(": ");
79         if(sta == 1) {
80             p(-1) pn
81             continue;
82         }
83         sort(a,a+n,cmp);
84         //f(0);
85         double end=-0x3f3f3f3f;//橫座標要很小....因為..你懂的
86         for(int i=0; i<n; i++) {
87             if(end<a[i].l) {
88                 end=a[i].r;
89                 count1++;
90             }
91         }
92 
93         p(count1) pn
94     }
95 
96 
97     return 0;
98 }