LightOj1388

题目链接：http://lightoj.com/volume_showproblem.php?problem=1388

题意：已知梯形的点A B的坐标，以及b c d的长度，求C D两点的坐标；默认AB是梯形平行边的较长边；

注意D点可能在A点的左边，AB可能不是水平的,剩下的求根据A点求D点，在根据D点求C点；

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<set>
using namespace std;
#define met(a, b) memset(a, b, sizeof(a))
#define maxn 10005
#define maxm 20005
#define INF 0x3f3f3f3f
typedef long long LL;

int main()
{
    int T, tcase = 1;
    scanf("%d", &T);
    while(T --)
    {
        double b, c, d, ax, ay, bx, by;

        scanf("%lf %lf %lf %lf %lf %lf %lf", &ax, &ay, &bx, &by, &b, &c, &d);

        double a = sqrt((ax-bx)*(ax-bx) + (ay-by)*(ay-by));///直线AB的距离；

        double sina = (by-ay) / a;
        double cosa = (bx-ax) / a;
        ///AB与水平线的sin和cos值；

        double p = (b+d+a-c)/2;
        double s = sqrt(p*(p-b)*(p-d)*(p-(a-c)));
        double h = s*2/(a-c);///D到AB的距离；

        int flag = 0;

        if(b*b > d*d + (a-c)*(a-c)) flag = 1;///角DAB为钝角;

        double t = sqrt(fabs(d*d - h*h));

        if(flag) t = -t;

        double dx = ax + (t*cosa - h*sina);
        double dy = ay + (h*cosa + t*sina);

        double cx = dx + c*cosa;
        double cy = dy + c*sina;

        printf("Case %d:
%.8lf %.8lf %.8lf %.8lf
", tcase++, cx, cy, dx, dy);
    }
    return 0;
}