Pizza Hub
Time Limit: 3000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 747 Accepted Submission(s): 161
Special Judge
Problem Description
Coffee Chicken has started a new restaurant named Pizza Hu...b! It provides various styles of pizzas, hamburgers, sandwiches, coffee, chickens and many other awesome Western cuisines. Welcome to Pizza Hub after this Multi-University Training Contest!
Since the pizzas are so exquisite, it is never a bad thing to design nice paper pads for them. The pizzas provided in Pizza Hub are sliced into triangles. The rectangular-shaped paper pads are cut from a paper strip of fixed width which is long enough. The pizza should be placed entirely on the pad; however, their borders are allowed to touch. Also, you are allowed to rotate the pizza.
As the customized paper strip is rather expensive, minimizing the size of the pizza pad can save a lot of money. Can you determine the minimum possible height of the pizza pad, given the width of the paper strip? The following picture illustrates the first sample test case.
Input
The first line of the input is a single integer T (1≤T≤50000), the number of test cases.
Each test case is a single line of seven integers x1,y1,x2,y2,x3,y3 (0≤x1,y1,x2,y2,x3,y3≤10000) and w (1≤w≤10000), where (x1,y1), (x2,y2) and (x3,y3)are Cartesian coordinates of the vertices of the pizza, and w is the width of the strip. It is guaranteed that the three vertices are not collinear.
Output
For each test case, display the minimum height of the pizza pad with an absolute or relative error of no more than 10−6. If it is impossible to make a pizza pad, display impossible instead.
Sample Input
2 0 0 3 0 0 4 10 0 0 3 0 0 4 1
Sample Output
2.400000000 impossible
Source
2018 Multi-University Training Contest 8
题意：给出三角形的三个坐标和矩形的宽，求出能够放进矩形的最小高度。
用向量来表示边的情况。
分一下情况：
向量a的那条边水平放置
向量a的那条边斜着放置
以下又分两种情况：向量b所在的那条边在向量a所在的那条边之上 ；
向量b所在的那条边在向量a所在的那条边之下 ；
#include<bits/stdc++.h>
using namespace std;
#define PI acos(-1.0)
typedef long long LL;
struct point
{
LL x,y;
} p[15];
LL operator *(point A,point B)
{
return A.x*B.x+A.y*B.y;
}
LL operator ^(point A,point B)
{
return A.x*B.y-B.x*A.y;
}
point operator-(point A,point B)
{
return (point)
{
A.x-B.x,A.y-B.y
};
}
double sum;
void js(point a,point b,LL w)
{
if(a*b<0)
return ;
if(a*a<=w*w)
{
if(a*b/sqrt(a*a)>w)
return ;
if((a^b)<0)
return ;
sum=min(sum,abs(a^b)/sqrt(a*a));
}
else
{
double h1=sqrt(a*a-w*w);
double t1=atan(h1/w);
double t2=PI/2-t1;
if((a^b)>=0)
{
double s1=acos(a*b/(sqrt(a*a)*sqrt(b*b)));
if(s1-t2>1e-9)
return ;
double L1=sqrt(b*b)*cos(s1+t1);
if(L1-w>0)
return ;
double h2=sqrt(b*b)*sin(s1+t1);
sum=min(sum,max(h1,h2));
}
else
{
double s1=acos(a*b/((sqrt(a*a)*sqrt(b*b))));
if(s1-t1>1e-9)
return ;
double L1=sqrt(b*b)*cos(t1-s1);
if(L1-w>0)
return ;
sum=min(sum,h1);
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
LL w;
for(int i=1; i<=3; i++)
{
scanf("%lld%lld",&p[i].x,&p[i].y);
}
scanf("%lld",&w);
sum=1e9+7;
js(p[3]-p[1],p[2]-p[1],w);
js(p[2]-p[1],p[3]-p[1],w);
js(p[3]-p[2],p[1]-p[2],w);
js(p[1]-p[2],p[3]-p[2],w);
js(p[1]-p[3],p[2]-p[3],w);
js(p[2]-p[3],p[1]-p[3],w);
for(int i=1; i<=3; i++)
{
p[i].y=-p[i].y;
}
js(p[3]-p[1],p[2]-p[1],w);
js(p[2]-p[1],p[3]-p[1],w);
js(p[3]-p[2],p[1]-p[2],w);
js(p[1]-p[2],p[3]-p[2],w);
js(p[1]-p[3],p[2]-p[3],w);
js(p[2]-p[3],p[1]-p[3],w);
if(sum>=1e9)
printf("impossible\n");
else
printf("%.9f\n",sum);
}
return 0;
}