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.
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
經典最近點對問題 //類似題目 POJ 3714 Raid (沒做)
參考 《算法導論》、《計算機算法設計與分析》(王曉東)
算導上的關於 證明P中至多有8個點可能處於該d x 2d 矩形區域中 有待理解
code:
[cpp]
#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)//前閉後開區間
{
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軸排序 果斷TLE Time Limit Exceeded 1007 5000MS 1812K
#2 按y軸排序 AC Accepted 1007 1125MS 1840K
*/
#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)//前閉後開區間
{
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軸排序 果斷TLE Time Limit Exceeded 1007 5000MS 1812K
#2 按y軸排序 AC Accepted 1007 1125MS 1840K
*/
