soj1209- 最短的距离(精度问题)

1209: 最短的距离

Time Limit: 1 Sec  Memory Limit: 128 MB Submit: 149  Solved: 5 [Submit][Status][Web Board]

Description

如图所示：

这里有一个平行于X轴的长方形（红色标出）和一个点(x,y)；你能帮UnkelTao计算下点到长方形的最短距离吗？

 

 

Input

多组测试数据；

每组数据含有10个实数，前八个分别表示长方形的四个坐标x0,y0,x1,y1,x2,y2,x3,y3;后两个表示点的坐标x,y;  其中每个数的绝对值均小于10000并且矩阵的四个坐标点均按顺时针或者逆时针给出。

保证输入图形为矩阵。.

 

Output

输出最短的距离值，结果保留2位小数。输出格式见样例

 

Sample Input

0 0 0 2 2 2 2 0 1 3
0 0 0 2 2 2 2 0 3 3

Sample Output

1.00
1.41

HINT

输入输出最好使用%lf

 

----------------------------------------------------

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<fstream>
using namespace std;
double eps=0.00000001;
double x1,x2,yy1,y2;
double x,y;
double sqr(double a)
{
    return a*a;
}
double minn(double a,double b,double c,double d)
{
    if (a-b<eps) b=a;
    if (b-c<eps) c=b;
    if (c-d<eps) d=c;
    return d;
}
double deal()
{
    double t,t1,t2,t3,t4;
        if (x-x1<=eps)
        {  t1=x-x1;t2=x2-x;
        t
            if (y-y2>=eps) return sqrt(sqr(x1-x)+sqr(y-y2));
            if (y-yy1<=eps) return sqrt(sqr(x1-x)+sqr(yy1-y));
            return x1-x+eps;
        }
        if (x-x2>=eps)
        {
            if (y-y2>=eps) return sqrt(sqr(x2-x)+sqr(y-y2));
            if (y-yy1<=eps) return sqrt(sqr(x2-x)+sqr(yy1-y));
            return x-x2+eps;
        }
        if (y-y2>=eps) return (y-y2+eps);
        if (y-yy1<=eps) return (yy1-y+eps);
      3=y-yy1;t4=y2-y;
        return minn(t1,t2,t3,t4)+eps;
}
int main()
{
    double t,a,b;
   // freopen("data.in","r",stdin);
  //  freopen("a.out","w",stdout);
    while (scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&yy1,&a,&b,&x2,&y2,&a,&b,&x,&y)!=EOF)
       {
           if (x1-x2>=eps) {t=x1;x1=x2;x2=t;}
           if (yy1-y2>=eps) {t=yy1;yy1=y2;y2=t;}
           printf("%.2lf\n",deal());
       }
}