title: woj1016 cherry blossom
date: 2020-03-18 20:00:00
categories: acm
tags: [acm,幾何,woj]

幾何題，判斷給出的點是否對稱

1 題目描述

March is wonderful in Wuhan University for the blooming cherry blossoms. Walking in the campus, you can smell the fragrance and feel the
romance with the tardiness falling of cherry blossom petals, and after a while you will see the ground will be covered by the beautiful
blossom petals.

Now, here comes the problem: the figure shown below is the cherry blossom petals on the ground, where the black points stand for blossom
petals. And this figure on the left is up-down symmetric as it is possible to cut the figure into two identical halves by the dashed line
shown in it, but the figure on the right is not.

This figure shows the sample input data

Given such a figure, can you tell me whether it is up-down symmetric or not. All the black points are different from each other.

2 輸入輸出

輸入格式
There are multiple test cases.For each test case, it contains:
Line 1: One integer N (1<=N<=1000) which specifies the number of the cherry blossom petals in this test case.
Line 2?N+1: Two integers X and Y (you are ensured that the absolute value of both the integers are less than) which specify the position of the
cherry blossom petal.
Input will be terminated by EndOfFile.

輸出格式
Print exactly one line for each test case. You should output YES if the figure is up-down symmetric, else output NO.

3 樣例

樣例輸入
5
0 0
2 0
1 1
0 2
2 2
4
0 0
2 0
1 1
0 2
樣例輸出
YES
NO

4 分析

//題意 輸入櫻花座標，判斷所有點是否上下對稱。
//因為數量小所以可以直接存圖 1000
//思路：找到縱座標最低最高點,得到對稱軸高度，然後判斷每個x座標上的點是否符合在對稱軸兩側
//輸入的時候做處理，使得所有的x座標相同的點都放到一個vector陣列，然後排序，那麼對應的i,len-i兩個點如果不對稱，就不對稱了。
//注意某個x座標奇數個點有一個點應該在對稱軸上

5 code

#include<iostream>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
using namespace std;

const double eps=1e-6;

vector<int>xcoor;
vector<int>ycoor[1005];
int a,b,num,len,maxh,minh;
double avg;
vector<int>::iterator iter;
bool flag;
int main(){
    while(cin>>num){
        flag=true;
        xcoor.clear();
        for(int i=0;i<num;i++)
            ycoor[i].clear();
        cin>>a>>b;
        minh=maxh=b;
        xcoor.push_back(a);
        ycoor[xcoor.size()-1].push_back(b);
        for(int i=1;i<num;i++){
            cin>>a>>b;
            minh=minh>b?b:minh;
            maxh=maxh>b?maxh:b;
            if((iter=find(xcoor.begin(),xcoor.end(),a))!=xcoor.end()){
                ycoor[iter-xcoor.begin()].push_back(b);
            }
            else{
                xcoor.push_back(a);
                ycoor[xcoor.size()-1].push_back(b);
            }
        }
        avg=((double)minh+maxh)/2;         //注意double
        for(int i=0;i<xcoor.size();i++){
                len=ycoor[i].size();
                sort(ycoor[i].begin(),ycoor[i].end());
                for(int j=0;j<len/2;j++)
                    if(fabs(ycoor[i][j]+ycoor[i][len-j-1]-2*avg-0)>eps) //兩邊的點是否對稱
                        {flag=false;break;}
                if(len&1==1){                          //一開始寫成^1了....
                    if(fabs(avg-ycoor[i][len/2])>eps) //中間點是否在對稱軸上
                        flag=false;
                }
                if(flag==false)
                    break;
        }
        if(flag)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}

title: woj1017 Billiard ball 幾何
date: 2020-03-18 21:43:00
categories: acm
tags: [acm,woj,幾何,數學]

一道三角幾何題，用到餘弦定理，圓周角定理，畫圖...

1 描述

There is a billiard table with acute triangle shape, and there are pockets at the three corners. A billiard ball hits the side of the table not at a
corner, it is reflected in the mirror direction. And it continues to travel on the table and we assume that the energy of this ball will not be
consumed, so the velocity of the ball is a constant number in the running process.

Now, you place the ball on any side of this billiard table and decide the slope of the line along which the ball starts at the origin. The trajectory
of the ball seems too complex to you, right? In order to simply this question, we will set restriction that the ball should bounce each side exactly
once and return to its start position at last. Under this restriction, can you find such trajectory that the time of ball running along it is shorter
than any other trajectories which satisfy this restriction?

2 輸入輸出

輸入格式
There are N test cases. The number of test cases N is given in the first line of the input file. For each test case, it contains six
integers (you are ensured that the absolute value of all the integers are less than) which specify the vertex point coordinates of the
acute triangle.

輸出格式
You should output the length of the shortest trajectory.The answer should be accurate to three fractional digits exactly one line for each test case.

3 樣例

樣例輸入
2
0 0 2 0 1 2
0 0 2 0 1 3
樣例輸出
3.200
3.600

4 分析

參考：
http://blog.sina.com.cn/s/blog_493bff030100041r.html

題目要求：
求小球經過兩次反射後，回到出發點的路徑的最小值。也就是三角形DEF周長的最小值。
題目要求小球和每邊碰撞且只碰一次，最後回到出發點

設小球路線在BC,AC,AB上的點為D,E,F
D關於AB,AC的對稱點為D',D''

根據對稱，我們可以知道
DF = D’F

DE = D”E

AD = AD’= AD”

那麼，求⊿DEF的周長可以轉化為求折線D’F + FE + ED”的長度

又由於兩個等腰三角形⊿ADD’和⊿ADD”,有下面的角度關係：

∠D’AD” = ∠D’AB + ∠BAD + ∠DAC + ∠CAD” = 2∠BAD + 2∠DAC = 2∠A

猜想：如果AD儘可能短，折線變成直線的話，那麼所求就是最短的了。（可以想到垂線是最短的）

下面是具體的證明過程：

（1）將D, E兩點視為定點，求 DF + EF 的最小值　　見圖2

找點D關於邊AB的對稱點D”,當F為 D”E與AB的交點時,DF + EF的值最小，

有　　∠DFB = ∠D”FB = ∠EFA

同理，將D, F兩點視為定點, 有　　∠CED = ∠AEF 將E, F兩點視為定點, 有　　∠BDF = ∠CDE

見圖3

A + y + z = B + y + x = C + x + z = 180°

而　A + B + C = 180°

解得：x = A, y = B, z = C

（2） 設⊿DEF為垂足三角形，那麼

因為AD, BE, CF為三高，B, D, E, A四點共圓，則有：∠CED = ∠B （角B+角BAD=90°=角CED+角BED，根據圓周角定理角BED=角BAD）

C, E, F, B四點共圓，則有： ∠AEF = ∠B

故：　∠CED =　∠AEF = ∠B

同理：∠BDF = ∠CDE = ∠A, ∠AFE = ∠BFD = ∠C

見圖4　

（3） 由(1)和(2)可知，⊿DEF為垂足三角形時，所得周長最短。

計算公式：

見圖5

等腰三角形AD’D”中，

D’D”　= 2 * AD * sinA ( 或2 * BE * sinB 或2 * CF * sinC )

AD = b * c * sinA / a //面積公式

故D’D”　= 2 * b * c * (1-cosA * cosA) / a; //cos可以由余弦定理求得

由上即可計算結果。

牛牛們的計算方法：
(1) D’D”= a * cosA + b * cosB + c * cosC ( gg, llyy, index等牛牛)

(2) D’D”= ( (a^2* b^2+b2 * c^2 + c^2 * a^2) – (a^4 + b^4 + c^4)/2.0 )/(abc) ( mathsoft, c0053，jfvp等牛牛)
(3) D’D”= 8 * area * area / ( abc) ( pear 等牛牛)

大家有興趣可以自己證明一下，好好領會牛牛們的超級思維。收穫肯定很大～～～

5 code

#include<iostream>
#include<cmath> //sqrt
using namespace std;

double leng(int a1,int b1,int a2,int b2){
    return sqrt(pow(a1-a2,2)+pow(b1-b2,2)); //pow
}

int main(){
    int a1,b1,a2,b2,a3,b3;
    double a,b,c,cosa,ans;
    int T;
    cin>>T;
    while(T){
        cin>>a1>>b1>>a2>>b2>>a3>>b3;
        a=leng(a1,b1,a2,b2);
        b=leng(a1,b1,a3,b3);
        c=leng(a2,b2,a3,b3);
        cosa=(b*b+c*c-a*a)/(2*b*c);
        ans=2*b*c*(1-cosa*cosa)/a;
        printf("%.3lf\n",ans);

        T--;
    }
return 0;
}