Quoit Design

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

   
   

    
    2
0 0
1 1
2
1 1
1 1
3
-1.5 0
0 0
0 1.5
0
   
   

   
   

    
    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;
}