joj2487

 2487: Central Avenue Road

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	65536K	743	169	Standard

There are a lot of trees in the park of Jilin University. We want to open up a road in this forest in a straight line from one tree to another. In order to view more scenery, the difference of both sides of the road is minimum (zero or one). If other tree is just on the road, it is not counted on both sides. Of course, we have a lot of ways to build this road. The best way is that the distance of two reference trees is shortest.

Input

The first line of each case is the number of trees N (2 <= N <=100). The next N lines have two integers that is the x and y coordinates of the i-th tree. The input file is terminated by N=0.

Output

For each case, output the distance of the two reference trees that the road obeys the above rules and the distance is shortest. Keep three digits after decimal point.

Sample Input

4
0 0
3 4
10 0
0 10
0

Sample Output

5.000

Problem Source: provided by skywind

#include<stdio.h>
#include<math.h>
int main()
{
    double  a[102][3];
    int  n;
    double temp;
    while(scanf("%d",&n),n)
    {
        temp=0xffffffff;
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&a[i][1],&a[i][2]);
        }
        for(int i=1;i<n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                double d=sqrt((a[i][1]-a[j][1])*(a[i][1]-a[j][1])+
                              (a[i][2]-a[j][2])*(a[i][2]-a[j][2]));
                int left=0;
                int right=0;
                if(d<temp)
                {
                    if(a[i][1]==a[j][1])
                    {
                        for(int t=1;t<=n;t++)
                        {
                            if(a[t][1]==a[i][1])continue;
                            if(a[t][1]>a[i][1])right++;
                            else if(a[t][1]<a[i][1]) left++;
                        }
                    }
                    else
                    {
                        double k=(a[i][2]-a[j][2])/(a[i][1]-a[j][1]);
                        double b=a[i][2]-a[i][1]*k;
                        for(int t=1;t<=n;t++)
                        {//注意精度
                            if(fabs(a[t][2]-(k*a[t][1]+b))<0.000002)continue;
                            if(a[t][2]>k*a[t][1]+b)right++;
                            else if(a[t][2]<k*a[t][1]+b)left++;
                        }
                    }
                    if(fabs(left-right)<=1)
                    temp=d;
                }
            }
        }
        printf("%.3lf\n",temp);
    }
    return 0;
}
//这个题虽然不难但是让我纠结了几个小时。首先是
//看了很久都没有理解题意。到后来才明白
//再就是精度没有处理好这里特别强调一下对于
//有关浮点数的计算一定要注意精度的选择，因为
//计算机对数的精确度是有要求的。不要把电脑真的当成了人脑。