CFF brush question 202006-1 linear classifier

topic:

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Input sample 1

Output sample 1

No
No
Yes

Show results
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Problem-solving ideas

1. The focus of this question

The main thing you need to know is how to judge whether the A and B points in the training data summary are separated, and what the judgment standards and methods are.

2. Problem-solving ideas

Judgment criteria : The coordinates of the substitution point in the equation, >0 is at the top and <0 is at the bottom.

Specific ideas :

First, when judging each (m) straight line, the category (A or B) and judgment result of the first point (above or below the line) are used as the initial value of the judgment result, and the categories and judgments of subsequent points (n) are If the result is inconsistent with the initial result, the segmentation fails, and if it fails, the judgment loop will be exited immediately.

For example, initially A is at the top and B is at the bottom. Subsequent points can only have A at the top and B at the bottom, otherwise the segmentation will fail.

Code display

/*CFF试题202006-1，试题名称：期线性分类器 Time：20210209*/
/*主要需要弄懂如何判断训练数据汇总的A、B两类点是否分开，判断标准和方法是怎样的*/
/*判断标准：方程代入点的坐标，>0在上方，<0在下方 
	首先每条（m）直线判别时，先将第一个点的类别（A or B）和判断结果（直线的上方 or 下方）
	作为判定结果的初始值，此后的点（n）的类别和判断的结果与初始不一致则分割失败，失败则立即退出此次判断循环
	比如初始A在上方，B在下方，后续的点也只能A在上方，B在下方，否则就分割失败*/
/*标准源码获取：清月学习社*/
#include <iostream>
#include<stdio.h>// 包含sort函数 
using namespace std; 

// 建立点的的结构体 
struct Node
{
    
    
    int x;		// 横坐标 
    int y;		// 纵坐标
    char type;	// 点的类别 
};


int main()
{
    
    
    Node node[1000];
	
    bool sign[20];
	// 初始假定完美分割，1-完美分割，0-分割失败 
    for(int i=0;i<20;i++)
    {
    
    
        sign[i] = true; 
    }
    
	
    int m, n;
    cin >> n >> m;// 点和方程个数 
    
    //输入数据，存到结构体数组中
    for(int i=0;i<n;i++)
    {
    
    
        cin >> node[i].x >> node[i].y >> node[i].type;
    }
      
    
    for(int i=0;i<m;i++)
    {
    
    
    	int flagA = -1;
		int flagB = -1;
				
		int Q1, Q2, Q3;// 方程系数 
        cin >> Q1 >> Q2 >> Q3; 
		for(int j=0;j<n;j++)
    	{
    
    
    		int ans = (Q1 + Q2*node[j].x + Q3*node[j].y) > 0 ? 1 : 0; //1直线上方,0直线下方

			if(node[j].type == 'A')// A 
			{
    
    
				if(flagA == -1)// 初始化点在直线的哪一边 
					flagA = ans;
			
				else if(flagA != ans)// 如果后续的不在同一侧，则分割失败 
				{
    
    
					sign[i] = false;// 分割失败
					break;			// 分割失败则不再继续判别，直接跳出此次循环 
				}
			}		
			else if(node[j].type == 'B')// B 
			{
    
    
				if(flagB == -1)// 初始化点在直线的哪一边 
					flagB = ans;
			
				else if(flagB != ans)// 如果后续的不在同一侧，则分割失败 
				{
    
    
					sign[i] = false;// 分割失败
					break;			// 分割失败则不再继续判别，直接跳出此次循环 
				}
			}												
		}
		
    }
    // 按顺序输出判别结果 
	for(int i=0;i<m;i++)
	{
    
    
		if(sign[i] == false)
			cout << "No" << endl;
		else
			cout << "Yes" << endl; 
	}
	 
    return 0;
}

Debugging results:

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