Project Euler 102:Triangle containment 包含原点的三角形

Triangle containment

Three distinct points are plotted at random on a Cartesian plane, for which -1000 ≤ x, y ≤ 1000, such that a triangle is formed.

Consider the following two triangles:

A(-340,495), B(-153,-910), C(835,-947)X(-175,41), Y(-421,-714), Z(574,-645)

It can be verified that triangle ABC contains the origin, whereas triangle XYZ does not.

Using triangles.txt (right click and ‘Save Link/Target As…’), a 27K text file containing the co-ordinates of one thousand “random” triangles, find the number of triangles for which the interior contains the origin.

NOTE: The first two examples in the file represent the triangles in the example given above.

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包含原点的三角形

从笛卡尔平面中随机选择三个不同的点，其坐标均满足-1000 ≤ x, y ≤ 1000，这三个点构成一个三角形。

考虑下面两个三角形：

A(-340,495), B(-153,-910), C(835,-947)X(-175,41), Y(-421,-714), Z(574,-645)

可以验证三角形ABC包含原点，而三角形XYZ不包含原点。

在27K的文本文件triangles.txt（右击并选择“目标另存为……”）中包含了一千个“随机”三角形的坐标，找出其中包含原点在其内部的三角形的数量。

注意：文件中的前两个三角形就是上述样例。

解题

考虑了一下原点在三角形内部的三角形，原点到两个边的夹角应该有两个钝角，但是发现结果是510，表示不对，我是根据余弦定理就得costheta 这里有问题在实际中可能出现多于90度的情况然而在我计算中，我不知道怎么判断。还有个问题就是不知道我的这个想法是否有问题，下面的程序是不对的，留在这里待更改。

package Level4;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;

public class PE0102{
    public static void run(){
        int count = 0;
        ArrayList<int[]> list = readData();
        for(int i=0;i<list.size();i++){
            int arr[] = list.get(i);
            if(iscontainmentTriangle(arr)){
                count+=1;
            }
            if(i<10)
            System.out.println(iscontainmentTriangle(arr));
        }
        System.out.println(count);

    }
    // 判断原点是否在三角形内部
    public static boolean iscontainmentTriangle(int[] arr){
        int count =0;
        for(int i=0;i<arr.length-1;i+=2){
            int x1 = arr[i];
            int y1 = arr[i+1];
            for(int j=i+2;j<arr.length-1;j+=2){
                int x2 = arr[j];
                int y2 = arr[j+1];
                if(isObtuseAngle(x1,y1,x2,y2))
                    count ++;
                    if(count ==2)
                    return true;
            }
        }
        return false;
    }

    // 是不是钝角
    public static boolean isObtuseAngle(int x1,int y1,int x2,int y2){
        long costheta = x1*y1 + x2*y2;
        if(costheta <0)
            return true;
        return false;
    }
    // 转换成整型数组
    public static int [] StringtoInt(String[] strArr){
        int[] IntArr = new int[strArr.length];
        for(int i=0;i<strArr.length;i++)
            IntArr[i] = Integer.parseInt(strArr[i]);
        return IntArr;
    }
    // 读取数据
    public static ArrayList<int[]> readData(){
        String filename= "src/Level4/p102_triangles.txt";
        ArrayList<int[]> list = new ArrayList<int[]>();
        try {
            BufferedReader bufferedReader = new BufferedReader(new FileReader(filename));
            String line = "";
            while((line=bufferedReader.readLine())!=null){
                String[] strArr = line.split(",");
                list.add(StringtoInt(strArr));
            }
        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return list;
    }
    public static void main(String[] args){
        long t0 = System.currentTimeMillis();
        run();
        long t1 = System.currentTimeMillis();
        long t = t1 - t0;
        System.out.println("running time="+t/1000+"s"+t%1000+"ms");

    }
}

Java Code

mathblog 中提到了根据三角形面积相等的方式求解，ABC = ABO + ACO +BCO

这里我们知道了三角形的三个点如何根据这三个点求面积，看了下面求解的方式，根据两个向量可以快速的求出向量所组成三角形的面积S= 向量交叉相乘差的绝对值的二分之一

wiki 中有说明

Java

package Level4;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;

public class PE0102{
    public static void run(){
        int count = 0;
        ArrayList<int[]> list = readData();
        for(int i=0;i<list.size();i++){
            int arr[] = list.get(i);
            if(iscontainmentTria(arr)){
                count+=1;
            }

        }
        System.out.println(count);
//        228
//        running time=0s16ms
    }
    // 判断原点是否在三角形内部
        public static boolean iscontainmentTria(int[] arr){
            int area = 0;
            int count =0;
            int X1 = arr[0] - arr[2];
            int Y1 = arr[1] - arr[3];
            int X2 = arr[4] - arr[2];
            int Y2 = arr[5] - arr[3];
            int area2 = area(X1,Y1,X2,Y2);
            for(int i=0;i<arr.length-1;i+=2){
                int x1 = arr[i];
                int y1 = arr[i+1];
                for(int j=i+2;j<arr.length-1;j+=2){
                    int x2 = arr[j];
                    int y2 = arr[j+1];
                    area +=area(x1,y1,x2,y2);
                }
            }
            if(area == area2)
                return true;
            return false;
        }
    // 这里面积的二倍
    public static int area(int X1,int Y1,int X2,int Y2){
        int area = Math.abs(X1*Y2 - X2*Y1);
        return area;
    }
    // 转换成整型数组
    public static int [] StringtoInt(String[] strArr){
        int[] IntArr = new int[strArr.length];
        for(int i=0;i<strArr.length;i++)
            IntArr[i] = Integer.parseInt(strArr[i]);
        return IntArr;
    }
    // 读取数据
    public static ArrayList<int[]> readData(){
        String filename= "src/Level4/p102_triangles.txt";
        ArrayList<int[]> list = new ArrayList<int[]>();
        try {
            BufferedReader bufferedReader = new BufferedReader(new FileReader(filename));
            String line = "";
            while((line=bufferedReader.readLine())!=null){
                String[] strArr = line.split(",");
                list.add(StringtoInt(strArr));
            }
        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return list;
    }
    public static void main(String[] args){
        long t0 = System.currentTimeMillis();
        run();
        long t1 = System.currentTimeMillis();
        long t = t1 - t0;
        System.out.println("running time="+t/1000+"s"+t%1000+"ms");
    }
}

Python

# coding=gbk

import time as time
import re
import math
import numpy as np
def run():
    filename = 'E:/java/projecteuler/src/Level4/p102_triangles.txt'
    mat = readData(filename)
    mat = np.array(mat)
    count = 0
    for line in mat:
        if isContainmentTraingle(line):
            count+=1
    print count 

def isContainmentTraingle(triangle):
    X1 = triangle[0] - triangle[2]
    Y1 = triangle[1] - triangle[3]
    X2 = triangle[4] - triangle[2]
    Y2 = triangle[5] - triangle[3]
    S = area(X1,Y1,X2,Y2)
    for i in range(0,4,2):
        for j in range(i+2,5,2):
            S -= area(triangle[i],triangle[i+1],triangle[j],triangle[j+1])
    return S == 0

def area(x1,y1,x2,y2):
    S = np.abs(x1*y2 - x2*y1)
    return S

def readData(filename):
    mat = list()
    file = open(filename)
    for line in file:
        row = line.split(',')
        row = [int(x) for x in row]
        mat.append(row)
    return mat 

t0 = time.time()
run()
t1 = time.time()
print "running time=",(t1-t0),"s"