Quoit Design

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42865 Accepted Submission(s): 11128

Problem Description

Have you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded.
In the field of Cyberground, the position of each toy is fixed, and the ring is carefully designed so it can only encircle one toy at a time. On the other hand, to make the game look more attractive, the ring is designed to have the largest radius. Given a configuration of the field, you are supposed to find the radius of such a ring.

Assume that all the toys are points on a plane. A point is encircled by the ring if the distance between the point and the center of the ring is strictly less than the radius of the ring. If two toys are placed at the same point, the radius of the ring is considered to be 0.

Input

The input consists of several test cases. For each case, the first line contains an integer N (2 <= N <= 100,000), the total number of toys in the field. Then N lines follow, each contains a pair of (x, y) which are the coordinates of a toy. The input is terminated by N = 0.

Output

For each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places.

Sample Input

2 0 0 1 1 2 1 1 1 1 3 -1.5 0 0 0 0 1.5 0

Sample Output

0.71 0.00 0.75

Author

CHEN, Yue

最近點對問題:

分治的思想，如果是大於3個點的時候，將所有的點按照x坐標（或者y坐標）排序，然后從中間線一份兩半，分別求出左邊的點的最近點距ldis，和右邊點的最近點距rdis，那么問題就是要將兩個部分的點合起來，令dis = min(rdis,ldis);那么中間如果想出現最近點對的時候必須要出現在中間線兩側距離為d的區域內，那么考慮區域內的左側的點，與其產生最近點的點一定在坐標的右側，那么為了不用枚舉區域內右邊所有的點，就要考慮對於每個點它如果要產生距離小於dis的對應點，肯定是在以它為圓心的dis為半徑的園內，那么這個圓覆蓋這個帶狀區域面積最大的情況就是圓心在分界線上的時候。如圖

注意：在處理中間帶的時候將帶狀區域內的點按照y坐標排序（如果之前是按照y排序的，則現在按照x坐標排序）那么因為右側的任意兩點之間的距離要大於等於dis所以要想極限情況，就是在這個圓區域內找等邊三角形，因為從上到下處理的點，所以當前點的上方的點不在考慮，由於排序的時候是左右兩邊的點一起排序的所以要在當前點的編號向后處理7個點，如上圖中划出的三個點，如果發現他們來自同一側則不再處理

下面是模板代碼：

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<cmath>
  5 #include<algorithm>
  6 using namespace std;
  7 
  8 #define N 100005
  9 const int L = -1;
 10 const int R = 1;
 11 struct Node{
 12     int id;
 13     double x;
 14     double y;
 15 };
 16 Node num[N],cp[N];
 17 
 18 //double Dis(Node a, Node b){
 19 //    return sqrt(pow((a.x - b.x),2.0)+pow((a.y-b.y),2.0));
 20 //}
 21 double Dis(Node a, Node b){
 22     double x = a.x-b.x, y = a.y-b.y;
 23     return sqrt(x*x+y*y);
 24 }
 25 
 26 
 27 bool cmpx(Node a, Node b)
 28 {
 29     if(a.x==b.x) return a.y<b.y;
 30     else return a.x < b.x;
 31 }
 32 bool cmpy(Node a, Node b)
 33 {
 34     if(a.y==b.y) return a.x<b.x;
 35     else return a.y<b.y;
 36 }
 37 double solve(int low, int high)
 38 {
 39     double dis;
 40     int sum = high - low;
 41     if(sum == 0){ return 0; }//只有一個數
 42     else if(sum == 1){//兩個數
 43         dis = Dis(num[low],num[high]);
 44     }
 45     else if(sum == 2){//三個數
 46         double tm1,tm2,tm3;
 47         tm1 = Dis(num[low],num[low+1]);
 48         tm2 = Dis(num[low+1],num[high]);
 49         tm3 = Dis(num[low],num[high]);
 50         dis = min(tm1,min(tm2,tm3));
 51     }
 52     else //大於三個數
 53     {
 54         double lmin,rmin,mmin;
 55         int mid = (low+high)/2;
 56         int p = 0;
 57         int i, j;
 58         lmin = solve(low,mid);
 59         rmin = solve(mid+1,high);
 60         dis = min(lmin,rmin);
 61         /**-----------------提出來會變快-----------------*/
 62         double ldis = num[mid].x-dis;
 63         double rdis = num[mid].x+dis;
 64         for( i = low; i <= mid; i++) /**-----小於等於，不能是小於，因為下面標記 L R 了*/
 65         {
 66             if(num[i].x >= ldis)
 67             {
 68                 cp[p].id = L;//標記為屬於左邊的部分
 69                 cp[p].x = num[i].x;
 70                 cp[p].y = num[i].y;
 71                 p++;
 72             }
 73         }
 74         for( ; i <= high; i++)
 75         {
 76             if(num[i].x <= rdis)
 77             {
 78                 cp[p].id = R;//標記為右邊的點
 79                 cp[p].x = num[i].x;
 80                 cp[p].y = num[i].y;
 81                 p++;
 82             }
 83         }
 84         sort(cp,cp+p,cmpy);
 85         for( i = 0; i < p; i++)
 86         {
 87             for( j = 1; (j <= 7)&&(i+j)<p; j++)
 88             {
 89                 if(cp[i].id != cp[i+j].id)//最小值可能出現在分界線不同的兩邊
 90                 {
 91                     mmin = Dis(cp[i],cp[i+j]);
 92                     if(mmin<dis)
 93                         dis = mmin;
 94                 }
 95             }
 96         }
 97     }
 98     return dis;
 99 }
100 int main()
101 {
102     int n;
103     while(~scanf("%d",&n)&&n!=0)
104     {
105         double result = 0;
106         for(int i = 0; i < n; i++)
107         {
108             num[i].id = 0;
109             scanf("%lf%lf",&num[i].x,&num[i].y);
110         }
111         sort(num,num+n,cmpx);
112         result = solve(0,n-1);
113         printf("%.2f\n",result/2);
114     }
115     return 0;
116 }

下面是一開始wa的代碼：

對應的錯誤在上面代碼中用用/** -------*標識

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<cmath>
  5 #include<algorithm>
  6 using namespace std;
  7 #define N 100005
  8 const int L = -1;
  9 const int R = 1;
 10 struct Node{
 11     int id;
 12     double x;
 13     double y;
 14 };
 15 Node num[N],cp[N];
 16 
 17 double Dis(Node a, Node b){
 18     return sqrt(pow((a.x - b.x),2.0)+pow((a.y-b.y),2.0));
 19 }
 20 
 21 bool cmpx(Node a, Node b)
 22 {
 23     if(a.x==b.x) return a.y<b.y;
 24     else return a.x < b.x;
 25 }
 26 bool cmpy(Node a, Node b)
 27 {
 28     if(a.y==b.y) return a.x<b.x;
 29     else return a.y<b.y;
 30 }
 31 
 32 double solve(int low, int high)
 33 {
 34     double dis;
 35     int sum = high - low;
 36     if(sum == 0){ return 0; }//只有一個數
 37     else if(sum == 1){//兩個數
 38         dis = Dis(num[low],num[high]);
 39     }
 40     else if(sum == 2){//三個數
 41         double tm1,tm2,tm3;
 42         tm1 = Dis(num[low],num[low+1]);
 43         tm2 = Dis(num[low+1],num[high]);
 44         tm3 = Dis(num[low],num[high]);
 45         dis = min(tm1,min(tm2,tm3));
 46     }
 47     else //大於三個數
 48     {
 49         double lmin,rmin,mmin;
 50         int mid = (low+high)/2;
 51         int p = 0;
 52         int i, j;
 53         lmin = solve(low,mid);
 54         rmin = solve(mid+1,high);
 55         dis = min(lmin,rmin);
 56         for( i = low; i < mid; i++)
 57         {
 58             double ldis = num[mid].x - dis;
 59             if(num[i].x >= ldis)
 60             {
 61                 cp[p].id = L;//標記為屬於左邊的部分
 62                 cp[p].x = num[i].x;
 63                 cp[p].y = num[i].y;
 64                 p++;
 65             }
 66         }
 67         for( ; i < high; i++)
 68         {
 69             double rdis = num[mid].x+dis;
 70             if(num[i].x <= rdis)
 71             {
 72                 cp[p].id = R;//標記為右邊的點
 73                 cp[p].x = num[i].x;
 74                 cp[p].y = num[i].y;
 75                 p++;
 76             }
 77         }
 78         sort(cp,cp+p,cmpy);
 79         for( i = 0; i < p; i++)
 80         {
 81             for( j = 1; (j <= 7)&&(i+j)<p; j++)
 82             {
 83                 if(cp[i].id != cp[i+j].id)//最小值可能出現在分界線不同的兩邊
 84                 {
 85                     mmin = Dis(cp[i],cp[i+j]);
 86                     if(mmin<dis)
 87                         dis = mmin;
 88                 }
 89             }
 90         }
 91     }
 92     return dis;
 93 }
 94 int main()
 95 {
 96     int n;
 97     while(~scanf("%d",&n)&&n!=0)
 98     {
 99         double result = 0;
100         for(int i = 0; i < n; i++)
101         {
102             num[i].id = 0;
103             scanf("%lf%lf",&num[i].x,&num[i].y);
104         }
105         sort(num,num+n,cmpx);
106         result = solve(0,n-1);
107         printf("%.2f\n",result/2);
108     }
109     return 0;
110 }

View Code

下面給出大神的模板代碼：（雖然內容差不多，感覺一下代碼風格）

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cmath>
 4 using namespace std;
 5 #define N 100007
 6 
 7 using namespace std;
 8 
 9 struct Point
10 {
11     double x,y;
12 }pt[N];
13 int a[N];
14 
15 int n;
16 
17 bool cmp(Point a, Point b)
18 {
19     if (a.x != b.x) return a.x < b.x;
20     else return a.y < b.y;
21 }
22 bool cmp_y(int id1, int id2){
23     return pt[id1].y < pt[id2].y;
24 }
25 double getDis(const Point &a, const Point &b)
26 {
27    double x = a.x - b.x;
28    double y = a.y - b.y;
29    return sqrt(x*x + y*y);
30 }
31 double solve(int l, int r)
32 {
33     double ans = 0;
34     if (r - l + 1 <= 3)
35     {
36         if (r - l + 1 == 1) return ans;
37         ans = getDis(pt[l], pt[l + 1]);
38         if (r - l + 1 == 2) return ans;
39         for (int i = l; i < r; ++i)
40         {
41             for (int j = i + 1; j <= r; ++j)
42             {
43                 ans = min(ans, getDis(pt[i],pt[j]));
44             }
45         }
46         return ans;
47     }
48     int m = (l + r) >> 1;
49     double s1 = solve(l, m);
50     double s2 = solve(m + 1,  r);
51     ans = min(s1,s2);
52     int k = 0;
53     for (int i = m - 1; i >= l && pt[m].x - pt[i].x <= ans; --i) a[k++] = i;
54     for (int i = m + 1; i <= r && pt[i].x - pt[m].x <= ans; ++i) a[k++] = i;
55     sort(a, a + k, cmp_y);
56     for (int i = 0; i < k; ++i)
57     {
58         for (int j = i + 1; j < k && j <= i + 7; ++j)
59         {
60             ans = min(ans, getDis(pt[a[i]], pt[a[j]]));
61         }
62     }
63     return ans;
64 }
65 int main()
66 {
67     while (~scanf("%d",&n))
68     {
69         if (!n) break;
70         for (int i = 0; i < n; ++i) scanf("%lf%lf",&pt[i].x, &pt[i].y);
71         sort(pt, pt + n, cmp);
72         printf("%.2lf\n",solve(0, n - 1)/2.0);
73     }
74     return 0;
75 }

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