http://www.lightoj.com/volume_showproblem.php?problem=1062

题意：问两条平行边间的距离，给出从同一水平面出发的两条相交线段长，及它们交点到水平面的高。

思路：计算几何怎么可能直接算出答案orz解了好久方程觉得不对，应该是二分枚举平行边的距离，通过相似三角形，算出交点的高，与题目比较，小于误差范围就行了。

/** @Date    : 2016-12-10-18.18  * @Author  : Lweleth (SoungEarlf@gmail.com)  * @Link    : https://github.com/  * @Version :  */#include<bits/stdc++.h>#define LL long long#define PII pair#define MP(x, y) make_pair((x),(y))#define fi first#define se second#define PB(x) push_back((x))#define MMG(x) memset((x), -1,sizeof(x))#define MMF(x) memset((x),0,sizeof(x))#define MMI(x) memset((x), INF, sizeof(x))using namespace std;const int INF = 0x3f3f3f3f;const int N = 1e5+20;const double eps = 1e-8;int main(){    int T;    int cnt = 0;    cin >> T;    while(T--)    {        double x, y, c;        scanf("%lf%lf%lf", &x ,&y ,&c);        double h1, h2, m;        double s = 0;        double l = 0, r = min(x, y);        while(r - l > eps)        {            m = (l + r) / 2.00;            //cout << m << " ";            h1 = sqrt(x * x - m * m);            h2 = sqrt(y * y - m * m);            s = h1 * h2 / (h1 + h2);            //cout << s << " ";            if(fabs(s - c) <= eps)                break;            else if(s < c)                r = m;            else if(s > c)                l = m;        }        printf("Case %d: %.8lf\n", ++cnt, m);    }    return 0;}

LightOJ 1062 - Crossed Ladders 基础计算几何