Quoit Design
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20332 Accepted Submission(s): 5216
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.
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.
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
经典最近点对问题 //类似题目 POJ 3714 Raid (没做)
参考 《算法导论》、《计算机算法设计与分析》(王晓东)
算导上的关于 证明P中至多有8个点可能处于该d x 2d 矩形区域中 有待理解
code:
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#define N 100002
#define INF 0x3f3f3f3f
using namespace std;
struct NODE{ double x, y;};
NODE edge[N], b[N];
int cmp_x(const void *a, const void *b)
{
NODE *aa = (NODE*) a;
NODE *bb = (NODE*) b;
return aa->x > bb->x ? 1: -1;
}
int cmp_y(const void *a, const void *b)
{
NODE *aa = (NODE*) a;
NODE *bb = (NODE*) b;
return aa->y > bb->y ? 1: -1;
}
inline double dis( NODE a, NODE b)
{
double xx =a.x - b.x, yy = a.y - b.y;
return sqrt(xx*xx+yy*yy);
}
double find(int l, int r)//前闭后开区间
{
if(r-l == 1) return INF;
if(r-l == 2) return dis(edge[l], edge[r-1]);
int i, j, k;
double dm, tmp;
int m = (l+r)/2;
double d1 = find(l, m);
double d2 = find(m, r);
dm = d1 < d2 ? d1:d2;
for(i=l,k=0;i<r;i++)
if(fabs(edge[m].x-edge[i].x)<=dm )
b[k++] = edge[i];
qsort(b,k,sizeof(b[0]),cmp_y);
for(i=0;i<k;i++)
for(j=i+1;j<k;j++)
{
tmp = dis(b[i],b[j]);
if(tmp<dm)dm=tmp;
}
return dm;
}
int main()
{
int n, i;
while(scanf("%d",&n),n)
{
for(i=0;i<n;i++)
// #1 scanf("%lf%lf",&edge[i].x, &edge[i].y);
/* #2 */scanf("%lf%lf",&edge[i].y, &edge[i].x);
qsort(edge,n,sizeof(edge[0]),cmp_x);
double d = find(0,n);
printf("%.2lf\n",d/2);
}
return 0;
}
/*
#1 按x轴排序 果断TLE Time Limit Exceeded 1007 5000MS 1812K
#2 按y轴排序 AC Accepted 1007 1125MS 1840K
*/