三分法

http://hi.baidu.com/czyuan_acm/blog/item/8cc45b1f30cefefde1fe0b7e.html

三分法模版

double Calc(Type a)
{
    /* 根据题目的意思计算 */
}

void Solve(void)
{
    double Left, Right;
    double mid, midmid;
    double mid_value, midmid_value;
    Left = MIN; Right = MAX;
    while (Left + EPS < Right)
    {
        mid = (Left + Right) / 2;
        midmid = (mid + Right) / 2;
        mid_area = Calc(mid);
        midmid_area = Calc(midmid);
        // 假设求解最大极值.
        if (mid_area >= midmid_area) Right = midmid;
        else Left = mid;
    }
}

 

 ZOJ 3203

如图，人左右走动，求影子L的最长长度。
根据图，很容易发现当灯，人的头部和墙角成一条直线时(假设此时人站在A点)，此时的长度是影子全在地上的最长长度。当人再向右走时，影子开始投影到墙上，当人贴着墙，影子长度即为人的高度。所以当人从A点走到墙，函数是先递增再递减，为凸性函数，所以我们可以用三分法来求解。

#include<iostream>
#include<cstring>
#include <cstdio>
#include<string>
#include<queue>
#include<vector>
#include <set>
#include<ctime>
#include<cmath>
#include <cstdlib>
#include<algorithm>
#include <iomanip>
#include <bitset>
using namespace std;

#define LL long long 
#define eps 10e-10
#define Max 110

double H,h,D;

double Calc(double x)
{
    return (h * D - H * x) / (D - x) + x;
}

double Solve()
{
    double Left, Right;
    double mid, midmid;
    double mid_value, midmid_value;
    Left = 0 ,Right = h/H*D;
    while (Left + eps < Right)
    {
        mid = (Left + Right) / 2;
        midmid = (mid + Right) / 2;
        mid_value = Calc(mid);
        midmid_value = Calc(midmid);
        // 假设求解最大极值.
        if (mid_value >= midmid_value) Right = midmid;
        else Left = mid;
    }
    return mid_value;
}
int main(){
    int T ; scanf("%d",&T);
    while(T--){
        scanf("%lf%lf%lf",&H,&h,&D);
        printf("%.3f\n",Solve());
    }
}