要使汽车能转过此弯道，那么就是汽车的左边尽量贴着那个直角点，而汽车的右下后方的点尽量贴着最下面的边。
我们以O点为原点建立直角坐标系，我们可以根据角a给出P点横坐标的函数F(a)
那么很容易得到：
其中有条件：，可以很容易证明是一个单峰函数，所以接下来就是三分了，如果的最大值小于等于
y，那么就能通过此直角弯道，否则就通不过。
代码如下：
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#define pi 3.1415
using namespace std;
const double eps = 1e-4;
double l,x,y,w;
double calu(double a){
return l*cos(a)+(w-x*cos(a))/sin(a);
}
double ternary_search(double l,double r){
double ll,rr;
while(r-l>eps){
ll=(2*l+r)/3;
rr=(2*r+l)/3;
if(calu(ll)>calu(rr))
r=rr;
else
l=ll;
}
return r;
}
int main()
{
while(cin>>x>>y>>l>>w){
double l=0,r=pi/2;
double tmp=ternary_search(l,r);
if(calu(tmp)<=y)
puts("yes");
else
puts("no");
}
return 0;
}
题目大意：给你一个拐角，两边的路的宽度分别为x，y。汽车的长和宽分别为h，w。问你这个汽车否转弯成功。
解题思路：如图，枚举角度。
这是一个凸函数所以三分枚举角度。
Turn the corner
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2009 Accepted Submission(s): 765
Problem Description
Mr. West bought a new car! So he is travelling around the city.
One day he comes to a vertical corner. The street he is currently in has a width x, the street he wants to turn to has a width y. The car has a length l and a width d.
Can Mr. West go across the corner?
Input
Every line has four real numbers, x, y, l and w.
Proceed to the end of file.
Output
If he can go across the corner, print "yes". Print "no" otherwise.
Sample Input
10 6 13.5 4
10 6 14.5 4
Sample Output
yes
no
Source
2008 Asia Harbin Regional Contest Online
[cpp]
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#include <iostream> #include<time.h> #include<stdio.h> #include<string.h> #include<stdlib.h> #include<string> #include<cmath> #include<map> #define eps 1e-10 #define PI acos(-1) using namespace std;
const int maxn = 220;
#define LL long long double x, y, l, w;
double Del(
double ang) {
double s = l*cos(ang)+w*sin(ang)-x;
double h = s*tan(ang)+w*cos(ang);
return h; }
int main() {
while(cin >>x>>y>>l>>w) {
double Max = 0.0;
double left = 0;
double right = PI/2.0;
while(right-left > eps) {
double mid = (right+left)/2.0;
double rmid = (mid+right)/2.0;
double xp1 = Del(mid);
double xp2 = Del(rmid);
if(xp1 > xp2) right = rmid;
else left = mid; Max = max(Max, max(xp1, xp2)); }
if(Max > y) cout<<
"no"<<endl;
else cout<<
"yes"<<endl; } }
以车长为玄长，画四分之一圆，以此圆心为圆心，此圆半径减去车宽为半径再画四分之一圆。如果路拐角小圆之内，为yes，否则为no，如图：
#include <iostream>
#include<time.h>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<cmath>
#include<map>
#define eps 1e-10
#define PI acos(-1)
using namespace std;
const int maxn = 220;
#define LL long long
double x, y, l, w;
double Del(double ang)
{
double s = l*cos(ang)+w*sin(ang)-x;
double h = s*tan(ang)+w*cos(ang);
return h;
}
int main()
{
while(cin >>x>>y>>l>>w)
{
double Max = 0.0;
double left = 0;
double right = PI/2.0;
while(right-left > eps)
{
double mid = (right+left)/2.0;
double rmid = (mid+right)/2.0;
double xp1 = Del(mid);
double xp2 = Del(rmid);
if(xp1 > xp2) right = rmid;
else left = mid;
Max = max(Max, max(xp1, xp2));
}
if(Max > y) cout<<"no"<<endl;
else cout<<"yes"<<endl;
}
}
如上图，这个问题就是一个矩形的一边的2个点分别固定在x和y轴上，该边和x轴的夹角θ在0到π/2的范围内滑动，看另一边所代表的直线y = -tanθ*x+w/cosθ+l*sinθ在θ滑动的过程中会不会滑到点（x,y）的上方 即求f(θ) = w/cosθ+l*sinθ-tanθ*x-y的最大值是否大于0 我的求法是对f(θ)求导，得f'(θ),再用二分法求f'(θ)=0的点θ1，判断f(θ1)是否大于0 使用该思路成功
AC......
