hdu1007 Quoit Design（最近点对）

最近点对

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25147    Accepted Submission(s): 6685

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
   
   
   
   

 

题目大意：给出一些点的坐标，然后求出最近的点对之间的距离，输出的结果是该距离的1/2；

解题思路：开始用暴力依次遍历每一个点对做了一遍提交，返回一个TLE。后来发现其实排序之后只要把每个点后面的十个点遍历一下就行，这样在很大程度上节省了时间。虽然这个方法看起来挺奇怪的。但是后来我在黑书《算法导论》第八章看到了这种方法，并且上面证明了这种方法的正确性。其实排序后只要遍历每个点之后的七个点就行了。还有其他的方法，太懒了，没有去写。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
struct point
{
    double x,y;
}p[100010];
bool cmp(point a,point b)
{
    if(a.x<b.x)
        return true;
    if(a.x==b.x)
        if(a.y<b.y)
            return true;
    return false;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0)break;
        double min=9999999;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        sort(p,p+n,cmp);
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<i+10&&j<n;j++)
            {
                double len;
                len=sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y));
                if(len<min)
                    min=len;
            }
        }
        printf("%.2lf\n",double(min/2.0));
    }
    return 0;
}

后来又用模板做了一遍（严谨一点，但是挺长的感觉）

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 100005;
const double MAX = 10e100, eps = 0.00001;
struct Point { double x, y; int index; };
Point a[N], b[N], c[N];
double closest(Point *, Point *, Point *, int, int);
double dis(Point, Point);
int cmp_x(const void *, const void*);
int cmp_y(const void *, const void*);
int merge(Point *, Point *, int, int, int);
inline double min(double, double);
int main(){
int n, i;
double d;
while (~scanf("%d", &n)) {
if(!n)break;
for (i = 0; i < n; i++)
scanf("%lf%lf", &(a[i].x), &(a[i].y));
qsort(a, n, sizeof(a[0]), cmp_x);
for (i = 0; i < n; i++)
a[i].index = i;
memcpy(b, a, n *sizeof(a[0]));
qsort(b, n, sizeof(b[0]), cmp_y);
d = closest(a, b, c, 0, n - 1);
printf("%.2lf\n", d/2.0);
}
return 0;
}
double closest(Point a[],Point b[],Point c[],int p,int q){
if (q - p == 1) return dis(a[p], a[q]);
if (q - p == 2) {
double x1 = dis(a[p], a[q]);
double x2 = dis(a[p + 1], a[q]);
double x3 = dis(a[p], a[p + 1]);
if (x1 < x2 && x1 < x3) return x1;
else if (x2 < x3) return x2;
else return x3;
}
int i, j, k, m = (p + q) / 2;
double d1, d2;
for (i = p, j = p, k = m + 1; i <= q; i++)
if (b[i].index <= m) c[j++] = b[i];
//数组c左半部保存划分后左部的点, 且对y是有序的.
else c[k++] = b[i];
d1 = closest(a, c, b, p, m);
d2 = closest(a, c, b, m + 1, q);
double dm = min(d1, d2);
//数组c左右部分分别是对y坐标有序的, 将其合并到b.
merge(b, c, p, m, q);
for (i = p, k = p; i <= q; i++)
if (fabs(b[i].x - b[m].x) < dm) c[k++] = b[i];
//找出离划分基准左右不超过dm的部分, 且仍然对y坐标有序.
for (i = p; i < k; i++)
for (j = i + 1; j < k && c[j].y - c[i].y < dm; j++){
double temp = dis(c[i], c[j]);
if (temp < dm) dm = temp;
}
return dm;
}
double dis(Point p, Point q){
double x1 = p.x - q.x, y1 = p.y - q.y;
return sqrt(x1 *x1 + y1 * y1);
}
int merge(Point p[], Point q[], int s, int m, int t){
int i, j, k;
for (i=s, j=m+1, k = s; i <= m && j <= t;) {
if (q[i].y > q[j].y) p[k++] = q[j], j++;
else p[k++] = q[i], i++;
}
while (i <= m) p[k++] = q[i++];
while (j <= t) p[k++] = q[j++];
memcpy(q + s, p + s, (t - s + 1) *sizeof(p[0]));
return 0;
}
int cmp_x(const void *p, const void *q){
double temp = ((Point*)p)->x - ((Point*)q)->x;
if (temp > 0) return 1;
else if (fabs(temp) < eps) return 0;
else return - 1;
}
int cmp_y(const void *p, const void *q){
double temp = ((Point*)p)->y - ((Point*)q)->y;
if (temp > 0) return 1;
else if (fabs(temp) < eps) return 0;
else return - 1;
}
inline double min(double p, double q)
{
return (p > q) ? (q): (p);
}