LightOJ 1062 - Crossed Ladders 基础计算几何

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 ([email protected])

  * @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;

}