HDU1007 Quoit Design 【分治】

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30505    Accepted Submission(s): 8017

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
    
    
    
    

题意：给定n个点，求距离最短的两点的距离的一半。

题解：開始用暴力法。结果超时。然后换成分治就过了，分治的过程是先将每一个点的坐标读入到数组里，再将数组依照x坐标排序，然后分治找最小值。递归终止条件是仅仅剩两个元素或三个元素，可是若仅依照x排序终于结果不一定是最小值，由于有可能左边的元素与右边的元素构成最小值，所以须要再依据y值进行一次排序，此时数据规模已经相当小了。能够用暴力直接求解。 

分治代码：

#include <stdio.h>
#include <math.h>
#include <algorithm>
#define maxn 100002
using std::sort;

struct Node{
	double x, y;
} arr[maxn], temp[maxn];

bool cmpx(Node a, Node b)
{
	return a.x < b.x;
}

bool cmpy(Node a, Node b)
{
	return a.y < b.y;
}

double calDist(int i, int j)
{
	double x = arr[i].x - arr[j].x;
	double y = arr[i].y - arr[j].y;
	return sqrt(x * x + y * y);
}

double divideAndConquer(int l, int r)
{
	if(r - l == 1) return calDist(l, r);
	else if(r - l == 2){
		double a = calDist(l, l + 1);
		double b = calDist(l + 1, r);
		double c = calDist(l, r);
		if(b > c) b = c;
		return a < b ? a : b;
	}
	int mid = (l + r) >> 1, i, j, id = 0;
	double a = divideAndConquer(l, mid);
	double b = divideAndConquer(mid + 1, r);
	double min = a < b ? a : b;
	for(i = l; i <= r; ++i)
		if(fabs(arr[i].x - arr[mid].x) < min) temp[id++] = arr[i];
	sort(temp, temp + id, cmpy);
	for(i = 0; i < id; ++i)
		for(j = i + 1; j < id; ++j){
			a = temp[j].y - temp[i].y;
			if(a >= min) break;
			b = temp[j].x - temp[i].x;
			a = sqrt(a * a + b * b);
			if(a < min) min = a;
		}
	return min;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	int n, i, j;
	double ans, x, y, len;
	while(scanf("%d", &n), n){
		for(i = 0; i < n; ++i)
			scanf("%lf%lf", &arr[i].x, &arr[i].y);
		sort(arr, arr + n, cmpx);
		printf("%.2lf\n", divideAndConquer(0, n - 1) / 2);
	}
	return 0;
}

 

 

原TLE代码：

#include <stdio.h>
#include <math.h>
#define maxn 100002

struct Node{
	double x, y;
} arr[maxn];

double cal(int i, int j)
{
	double x = arr[i].x - arr[j].x;
	double y = arr[i].y - arr[j].y;
	return sqrt(x * x + y * y);
}

int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	int n, i, j;
	double ans, x, y, len;
	while(scanf("%d", &n), n){
		for(i = 0, ans = -1; i < n; ++i){
			scanf("%lf%lf", &arr[i].x, &arr[i].y);
			for(j = 0; j < i; ++j){
				len = cal(i, j);
				if(len < ans || ans < 0) ans = len;
			}
		}
		printf("%.2lf\n", ans / 2);
	}
	return 0;
}