LightOJ 1062 - Crossed Ladders 基础计算几何

Posted 2020-08-22 ( m Lweleth)

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;
}