三分题目

zoj 3203

题意：求人从左到右走影子的最大长度L。

设人离开灯的长度为x,当灯、人的头顶和墙角在一条直线上时(此时人在A点)，影子全部投在地上，此时x = (H-h)*D/H;当人继续向前走，一部分影子就会投在墙上，当人走到最右边，影子长度即为人的高度。所以从A点到墙角，人的影子长度先增大后减小，是一个凸性函数。根据两个相似三角形把L表示成x的函数L= D+H-x-(H-h)*D/x,然后三分求解最大值。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#define LL long long
#define _LL __int64
#define eps 1e-9

using namespace std;

double H,h,D;
double cal(double mid)
{
    return D+H-mid-(H-h)*D/mid;
}

double solve(double low, double high)
{
    while(high - low > eps)
    {
        double mid = (low + high) / 2.0;
        double midmid = (mid + high) / 2.0;

        double ans1 = cal(mid);
        double ans2 = cal(midmid);
        if(ans1 < ans2)
            low = mid;
        else high = midmid;
    }
    return low;
}

int main()
{
    int test;
    scanf("%d",&test);
    while(test--)
    {
        scanf("%lf %lf %lf",&H,&h,&D);
        double ans = solve((H-h)*D/H,D);
        printf("%.3lf\n",cal(ans));
    }
    return 0;
}

hdu 4355

题意很简单，求出一个点x使其他所有点与该点的距离差s^3*wi最小。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#define LL long long
#define _LL __int64
#define eps 1e-8
#define PI acos(-1.0)

using namespace std;
const int INF = 0x3f3f3f3f;

struct node
{
    double x;
    double w;
}p[50010];
int n;

double cal(double x)
{
    double ans = 0;
    for(int i = 0; i < n; i++)
    {
        double tmp = fabs(p[i].x - x);
        ans += tmp * tmp * tmp * p[i].w;
    }
    return ans;
}

double solve(double low, double high)
{
    double ans1,ans2;
    while(high - low > eps)
    {
        double mid = (low+high)/2.0;
        double midmid = (mid + high)/2.0;

         ans1 = cal(mid);
         ans2 = cal(midmid);

        if(ans1 < ans2)
            high = midmid;
        else
            low = mid;
    }
    return low;
}

int main()
{
    int test;
    scanf("%d",&test);
    double low, high;

    for(int item = 1; item <= test; item++)
    {
        scanf("%d",&n);
        low = INF;
        high = -INF;
        for(int i = 0; i < n; i++)
        {
            scanf("%lf %lf",&p[i].x,&p[i].w);
            low = min(low, p[i].x);
            high = max(high,p[i].x);
        }

        double ans = solve(low, high);
        printf("Case #%d: %.0lf\n",item,cal(ans));
    }
    return 0;
}

hdu 2438

题意：给出一个直角弯道，弯道的宽分别为x和y，汽车的长和宽为l和d，问汽车能否通过弯道。

思路：

汽车要通过弯道，其左边要尽量靠着O点，右下角要尽量贴着地面。如图所示，那么在这样的情况下，汽车能通过的条件是右上角Q点到O点所在直线的距离要小于等于路的宽度y。设角度a如图，那么把要求变量表示成角度的函数 f(a) = l*cos（a） -(x - d/cos(a) ) / tan(a) 。即f(a) = l*cos(a) - (x*sin(a)-d)/sin(a)。三分法就f(a)的最大值与y比较。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#define LL long long
#define _LL __int64
#define eps 1e-8
#define PI acos(-1.0)

using namespace std;
const int INF = 0x3f3f3f3f;

double x,y,l,w;
double cal(double t)
{
     return l*cos(t) + (w-x*cos(t))/sin(t);
}

double solve(double low, double high)
{
    while(high - low > eps)
    {

        double mid= (high - low)/3+low;
        double midmid= (high - low)*2/3+low;

        double ans1 = cal(mid);
        double ans2 = cal(midmid);

        if(ans1 < ans2)
            low = mid;
        else
            high = midmid;
     }
     return low;
}

int main()
{
    while(~scanf("%lf %lf %lf %lf",&x,&y,&l,&w))
    {
        double low = 0;
        double high = acos(-1.0)/2;
        double ans = solve(low, high);

        //printf("%lf\n",ans);

        if(cal(ans) <= y)
            printf("yes\n");
        else printf("no\n");
    }
    return 0;
}

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