Quoit Design
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
Source
ZJCPC2004
Recommend
JGShining
說實話，這篇文章什麼意思，我看了半天都沒有看懂！不過這不影響做題，這道題實際上就是要求一堆坐標裡最短的坐標之間的距離的1/2.
具體算法在《編程之美》中講得很詳細！
代碼如下：
[cpp]
/*
*最近點對的問題
*/
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int SIZE = 100005;
const int L = -1;
const int R = 1;
typedef struct
{
int index;
double x;
double y; /*用於記錄坐標點*/
}coord;
coord num[SIZE], c[SIZE]/*用作輔助數組*/;
double getDistance(coord &bi1, coord &bi2) /*求得兩點之間的距離*/
{
return sqrt(pow(bi1.x - bi2.x, 2.0) + pow(bi1.y - bi2.y, 2.0));
}
bool cmpx(coord &bi1, coord &bi2)
{
if (bi1.x == bi1.x)
return bi1.y < bi2.y;
else
return bi1.x < bi2.x;
}
bool cmpy(coord &bi1, coord &bi2)
{
if (bi1.y == bi2.y)
return bi1.x < bi2.x;
else
return bi1.y < bi2.y;
}
inline double min(double &bi1, double &bi2, double &bi3)
{
double minLength;
minLength = bi1 > bi2 ? bi2 : bi1;
minLength = minLength > bi3 ? bi3 : minLength;
return minLength;
}
inline double minDist(double &bi1, double &bi2)
{
if (bi1 > bi2)
return bi2;
return bi1;
}
double divide_conquer(int low, int high) /*分治法求最小距離*/
{
double dis;
int count = high - low;
if (count == 0)
{
return 0;
}
else if (count == 1) /*兩個數*/
{
dis = getDistance(num[low], num[high]);
}
else if (count == 2) /*三個數*/
{
double temp1, temp2, temp3;
temp1 = getDistance(num[low], num[low + 1]);
temp2 = getDistance(num[low + 1], num[high]);
temp3 = getDistance(num[low], num[high]);
dis = min(temp1, temp2, temp3);
}
else /*大於三個數的情況*/
{
double leftmin, rightmin, min;
int mid = (low + high) / 2;
int p = 0;
int i, j;
leftmin = divide_conquer(low, mid); /*求得左邊部分的最小值*/
rightmin = divide_conquer(mid + 1, high); /*求得右邊部分的最小值*/
dis = minDist(leftmin, rightmin);
/*下面從所有坐標點中找出所有x在leftCoord到rightCoord之間的點*/
for (i = low; i <= mid; i++)
{
double leftCoord = num[mid].x - dis;
if (num[i].x >= leftCoord)
{
c[p].index = L; /*標識屬於左邊部分*/
c[p].x = num[i].x;
c[p].y = num[i].y;
p++;
}
}
for ( ; i <= high; i++)
{
double rightCoord = num[mid].x + dis;
if (num[i].x <= rightCoord)
{
c[p].index = R; /*標識屬於右邊部分*/
c[p].x = num[i].x;
c[p].y = num[i].y;
p++;
}
}
sort(c, c + p, cmpy); /*找到的點再從小到大按照y排序一次*/
for (i = 0; i < p; i++)
{
if (c[i].index == L) /*左邊的點一個一個地搜索,按照規律,我們只要搜索之後的7個點就可以了*/
{
for (j = 1; (j <= 7) && (i + j < p); j++)
{
if (c[i + j].index == R) /*這個點還必須屬於右邊*/
{
min = getDistance(c[i], c[i + j]);
if(min < dis)
{
dis = min;
}
}
}
}
}
}
return dis;
}
int main ()
{
int n;
while (cin >> n && n != 0)
{
double result = 0;
for (int i = 0; i < n; i++)
{
num[i].index = 0;
cin >> num[i].x >> num[i].y;
}
sort (num, num + n, cmpx);
result = divide_conquer(0, n - 1);
printf("%.2lf\n", result / 2);
}
//system ("pause");
return 0;
}
/*
*最近點對的問題
*/
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int SIZE = 100005;
const int L = -1;
const int R = 1;
typedef struct
{
int index;
double x;
double y; /*用於記錄坐標點*/
}coord;
coord num[SIZE], c[SIZE]/*用作輔助數組*/;
double getDistance(coord &bi1, coord &bi2) /*求得兩點之間的距離*/
{
return sqrt(pow(bi1.x - bi2.x, 2.0) + pow(bi1.y - bi2.y, 2.0));
}
bool cmpx(coord &bi1, coord &bi2)
{
if (bi1.x == bi1.x)
return bi1.y < bi2.y;
else
return bi1.x < bi2.x;
}
bool cmpy(coord &bi1, coord &bi2)
{
if (bi1.y == bi2.y)
return bi1.x < bi2.x;
else
return bi1.y < bi2.y;
}
inline double min(double &bi1, double &bi2, double &bi3)
{
double minLength;
minLength = bi1 > bi2 ? bi2 : bi1;
minLength = minLength > bi3 ? bi3 : minLength;
return minLength;
}
inline double minDist(double &bi1, double &bi2)
{
if (bi1 > bi2)
return bi2;
return bi1;
}
double divide_conquer(int low, int high) /*分治法求最小距離*/
{
double dis;
int count = high - low;
if (count == 0)
{
return 0;
}
else if (count == 1) /*兩個數*/
{
dis = getDistance(num[low], num[high]);
}
else if (count == 2) /*三個數*/
{
double temp1, temp2, temp3;
temp1 = getDistance(num[low], num[low + 1]);
temp2 = getDistance(num[low + 1], num[high]);
temp3 = getDistance(num[low], num[high]);
dis = min(temp1, temp2, temp3);
}
else /*大於三個數的情況*/
{
double leftmin, rightmin, min;
int mid = (low + high) / 2;
int p = 0;
int i, j;
leftmin = divide_conquer(low, mid); /*求得左邊部分的最小值*/
rightmin = divide_conquer(mid + 1, high); /*求得右邊部分的最小值*/
dis = minDist(leftmin, rightmin);
/*下面從所有坐標點中找出所有x在leftCoord到rightCoord之間的點*/
for (i = low; i <= mid; i++)
{
double leftCoord = num[mid].x - dis;
if (num[i].x >= leftCoord)
{
c[p].index = L; /*標識屬於左邊部分*/
c[p].x = num[i].x;
c[p].y = num[i].y;
p++;
}
}
for ( ; i <= high; i++)
{
double rightCoord = num[mid].x + dis;
if (num[i].x <= rightCoord)
{
c[p].index = R; /*標識屬於右邊部分*/
c[p].x = num[i].x;
c[p].y = num[i].y;
p++;
}
}
sort(c, c + p, cmpy); /*找到的點再從小到大按照y排序一次*/
for (i = 0; i < p; i++)
{
if (c[i].index == L) /*左邊的點一個一個地搜索,按照規律,我們只要搜索之後的7個點就可以了*/
{
for (j = 1; (j <= 7) && (i + j < p); j++)
{
if (c[i + j].index == R) /*這個點還必須屬於右邊*/
{
min = getDistance(c[i], c[i + j]);
if(min < dis)
{
dis = min;
}
}
}
}
}
}
return dis;
}
int main ()
{
int n;
while (cin >> n && n != 0)
{
double result = 0;
for (int i = 0; i < n; i++)
{
num[i].index = 0;
cin >> num[i].x >> num[i].y;
}
sort (num, num + n, cmpx);
result = divide_conquer(0, n - 1);
printf("%.2lf\n", result / 2);
}
//system ("pause");
return 0;
} 上面的那段代碼，說實話，還有很大的問題，不過在杭電上居然也通過了，可見杭電數據之弱！現在發一段修改了bug的代碼！這一段代碼沒有錯誤！
[cpp] view plaincopyprint?
/*
*最近點對的問題
*/
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int SIZE = 100005;
const int L = -1;
const int R = 1;
typedef struct
{
int index;
double x;
double y; /*用於記錄坐標點*/
}coord;
coord num[SIZE], c[SIZE]/*用作輔助數組*/;
double getDistance(coord &bi1, coord &bi2) /*求得兩點之間的距離*/
{
return sqrt(pow(bi1.x - bi2.x, 2.0) + pow(bi1.y - bi2.y, 2.0));
}
bool cmpx(coord &bi1, coord &bi2)
{
if (bi1.x == bi1.x)
return bi1.y < bi2.y;
else
return bi1.x < bi2.x;
}
bool cmpy(coord &bi1, coord &bi2)
{
if (bi1.y == bi2.y)
return bi1.x < bi2.x;
else
return bi1.y < bi2.y;
}
inline double min(double &bi1, double &bi2, double &bi3)
{
double minLength;
minLength = bi1 > bi2 ? bi2 : bi1;
minLength = minLength > bi3 ? bi3 : minLength;
return minLength;
}
inline double minDist(double &bi1, double &bi2)
{
if (bi1 > bi2)
return bi2;
return bi1;
}
double divide_conquer(int low, int high) /*分治法求最小距離*/
{
double dis;
int count = high - low;
if (count == 0)
{
return 0;
}
else if (count == 1) /*兩個數*/
{
dis = getDistance(num[low], num[high]);
}
else if (count == 2) /*三個數*/
{
double temp1, temp2, temp3;
temp1 = getDistance(num[low], num[low + 1]);
temp2 = getDistance(num[low + 1], num[high]);
temp3 = getDistance(num[low], num[high]);
dis = min(temp1, temp2, temp3);
}
else /*大於三個數的情況*/
{
double leftmin, rightmin, min;
int mid = (low + high) / 2;
int p = 0;
int i, j;
leftmin = divide_conquer(low, mid); /*求得左邊部分的最小值*/
rightmin = divide_conquer(mid + 1, high); /*求得右邊部分的最小值*/
dis = minDist(leftmin, rightmin);
/*下面從所有坐標點中找出所有x在leftCoord到rightCoord之間的點*/
for (i = low; i <= mid; i++)
{
double leftCoord = num[mid].x - dis;
if (num[i].x >= leftCoord)
{
c[p].index = L; /*標識屬於左邊部分*/
c[p].x = num[i].x;
c[p].y = num[i].y;
p++;
}
}
for ( ; i <= high; i++)
{
double rightCoord = num[mid].x + dis;
if (num[i].x <= rightCoord)
{
c[p].index = R; /*標識屬於右邊部分*/
c[p].x = num[i].x;
c[p].y = num[i].y;
p++;
}
}
sort(c, c + p, cmpy); /*找到的點再從小到大按照y排序一次*/
for (i = 0; i < p; i++)
{
/*錯誤出現在這裡，上面我是只搜索了左邊，並且只計算了7個y值比c[i].y大的點到c[i]的距離,
可是實際上y值比c[i].y小的點也有可能與c[i]取得最小值，所以說上面的程序有錯誤。真正正確
的解答如下，那就是要搜索所有的點,並計算7個y值比c[i].y大的點到c[i]的距離,由於距離是兩個
點之間產生的,一個點的y值比另一個點小,那麼必然有另一個點的y值比一個點的大，由於這種關系,
從而保證了搜索出來的是最小的距離！
*/
for (j = 1; (j <= 7) && (i + j < p); j++)
{
if (c[i].index != c[i + j].index) /*最小值只可能出現在兩個分別屬於不同的邊的點上*/
{
min = getDistance(c[i], c[i + j]);
if(min < dis)
dis = min;
}
}
}
}
return dis;
}
int main ()
{
int n;
while (cin >> n && n != 0)
{
double result = 0;
for (int i = 0; i < n; i++)
{
num[i].index = 0;
cin >> num[i].x >> num[i].y;
}
sort (num, num + n, cmpx);
result = divide_conquer(0, n - 1);
printf("%.2lf\n", result / 2);
}
//system ("pause");
return 0;
}
/*
*最近點對的問題
*/
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int SIZE = 100005;
const int L = -1;
const int R = 1;
typedef struct
{
int index;
double x;
double y; /*用於記錄坐標點*/
}coord;
coord num[SIZE], c[SIZE]/*用作輔助數組*/;
double getDistance(coord &bi1, coord &bi2) /*求得兩點之間的距離*/
{
return sqrt(pow(bi1.x - bi2.x, 2.0) + pow(bi1.y - bi2.y, 2.0));
}
bool cmpx(coord &bi1, coord &bi2)
{
if (bi1.x == bi1.x)
return bi1.y < bi2.y;
else
return bi1.x < bi2.x;
}
bool cmpy(coord &bi1, coord &bi2)
{
if (bi1.y == bi2.y)
return bi1.x < bi2.x;
else
return bi1.y < bi2.y;
}
inline double min(double &bi1, double &bi2, double &bi3)
{
double minLength;
minLength = bi1 > bi2 ? bi2 : bi1;
minLength = minLength > bi3 ? bi3 : minLength;
return minLength;
}
inline double minDist(double &bi1, double &bi2)
{
if (bi1 > bi2)
return bi2;
return bi1;
}
double divide_conquer(int low, int high) /*分治法求最小距離*/
{
double dis;
int count = high - low;
if (count == 0)
{
return 0;
}
else if (count == 1) /*兩個數*/
{
dis = getDistance(num[low], num[high]);
}
else if (count == 2) /*三個數*/
{
double temp1, temp2, temp3;
temp1 = getDistance(num[low], num[low + 1]);
temp2 = getDistance(num[low + 1], num[high]);
temp3 = getDistance(num[low], num[high]);
dis = min(temp1, temp2, temp3);
}
else /*大於三個數的情況*/
{
double leftmin, rightmin, min;
int mid = (low + high) / 2;
int p = 0;
int i, j;
leftmin = divide_conquer(low, mid); /*求得左邊部分的最小值*/
rightmin = divide_conquer(mid + 1, high); /*求得右邊部分的最小值*/
dis = minDist(leftmin, rightmin);
/*下面從所有坐標點中找出所有x在leftCoord到rightCoord之間的點*/
for (i = low; i <= mid; i++)
{
double leftCoord = num[mid].x - dis;
if (num[i].x >= leftCoord)
{
c[p].index = L; /*標識屬於左邊部分*/
c[p].x = num[i].x;
c[p].y = num[i].y;
p++;
}
}
for ( ; i <= high; i++)
{
double rightCoord = num[mid].x + dis;
if (num[i].x <= rightCoord)
{
c[p].index = R; /*標識屬於右邊部分*/
c[p].x = num[i].x;
c[p].y = num[i].y;
p++;
}
}
sort(c, c + p, cmpy); /*找到的點再從小到大按照y排序一次*/
for (i = 0; i < p; i++)
{
/*錯誤出現在這裡，上面我是只搜索了左邊，並且只計算了7個y值比c[i].y大的點到c[i]的距離,
可是實際上y值比c[i].y小的點也有可能與c[i]取得最小值，所以說上面的程序有錯誤。真正正確
的解答如下，那就是要搜索所有的點,並計算7個y值比c[i].y大的點到c[i]的距離,由於距離是兩個
點之間產生的,一個點的y值比另一個點小,那麼必然有另一個點的y值比一個點的大，由於這種關系,
從而保證了搜索出來的是最小的距離！
*/
for (j = 1; (j <= 7) && (i + j < p); j++)
{
if (c[i].index != c[i + j].index) /*最小值只可能出現在兩個分別屬於不同的邊的點上*/
{
min = getDistance(c[i], c[i + j]);
if(min < dis)
dis = min;
}
}
}
}
return dis;
}
int main ()
{
int n;
while (cin >> n && n != 0)
{
double result = 0;
for (int i = 0; i < n; i++)
{
num[i].index = 0;
cin >> num[i].x >> num[i].y;
}
sort (num, num + n, cmpx);
result = divide_conquer(0, n - 1);
printf("%.2lf\n", result / 2);
}
//system ("pause");
return 0;
}
