HDU

问题描述:

 I'm sure this problem will fit you as long as you didn't sleep in your High School Math classes. 
Yes,it just need a little math knowledge and I know girls are always smarter than we expeted. 
So don't hesitate any more,come and AC it! 
Tell you three point A,B,C in a 2-D plain,then a line L with a slope K pass through C,you are going to find 
a point P on L,that makes |AP| + |PB| minimal. 

Input

The first line contain a t. 
Then t cases followed,each case has two parts,the first part is a real number K,indicating the slope,and the second 
part are three pairs of integers Ax,Ay,Bx,By,Cx,Cy(0 <=|Ax|,|Ay|,|Bx|,|By|,|Cx|,|Cy| <= 10000 ).

Output

Just out put the minimal |AP| + |PB|(accurate to two places of decimals ).

Sample Input

12.558467 6334 6500 9169 5724 1478

Sample Output

3450.55

问题描述:题目给我们一个斜率K，三个点(A,B,C)的坐标，让我们在一条直线(斜率为K,过C点),找到一点P使得|AP|+|BP|最小。

题目分析:我们先求出直线L的方程，然后根据直线方程来判断A与B是否同侧(判断方法就是：将点A,B代入直线方程中算它们的结果的积如果小于0，异测，等于0，至少有一点在直线上，大于0，同侧，判断原理就是：点在直线上方结果大于0，下方小于0，中等于0）;

如果异侧：则直接求A与B的距离

如果同侧：则求出A点（B点也可）关于直线L对称的点D，求出B与D 的距离。（求D点时，可以先求出B与D的中点mid（联立方程即可))

代码如下:

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>using namespace std;struct Point{    double x,y;};double get_dis(Point a,Point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int main(){    int t;    scanf("%d",&t);    while (t--) {        double k;        Point a,b,c;        scanf("%lf",&k);        scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y);        double A=k*a.x+c.y-k*c.x-a.y;        double B=k*b.x+c.y-k*c.x-b.y;        if (A*B<=0) {            printf("%.2f\n",get_dis(a,b));        }        else {            if (k==0) {//当k等于0的时候，中点公式                Point d;                d.x=a.x;                d.y=2*c.y-a.y;                printf("%.2f\n",get_dis(b,d));            }            else {                double k1,k2,b1,b2;                Point mid,d;                k1=k,k2=-1.0/k;                b1=c.y-k*c.x;                b2=a.y+a.x/k;                mid.x=(b2-b1)/(k1-k2);//中点                mid.y=k2*mid.x+b2;                d.x=2*mid.x-a.x;                d.y=2*mid.y-a.y;                printf("%.2f\n",get_dis(b,d));            }        }    }    return 0;}