题目

思路

判定点集合是否都在同一侧，可以将点代入解析式，如果都>0或都<0则在同一侧。

坑

….逻辑不难，但TM的要注意，标准输出是‘Yes’，不是’YES’！！害得我一直错！！

代码

#include<iostream>
#include<vector>
#include<string.h>
using namespace std;
struct point{
    int x,y;
    char type;
}; 
struct line{
    int theta0,theta1,theta2;
};
vector<point> points;
line lines[21];
int count_A=0,count_B=0;
int main(){
    memset(lines,0,sizeof(line));//初始化数组
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++){
            point temp_point;
        cin>>temp_point.x>>temp_point.y>>temp_point.type;
        if(temp_point.type =='A'){
            count_A++;
        }else{
            count_B++;
        }
        points.push_back(temp_point);
    } 
    for(int j=0;j<m;j++){
        line temp_line;
        cin>>temp_line.theta0>>temp_line.theta1>>temp_line.theta2;
        lines[j] =temp_line;
    }
    for(int j=0;j<m;j++){
        int A1=0,A0=0,B1=0,B0=0;
        for(int i=0;i<n;i++){
            int result =lines[j].theta0+lines[j].theta1*points[i].x+lines[j].theta2*points[i].y;            
            if(points[i].type =='A'){
                if(result>=0){
                    A1++;
                }else{
                    A0++;
                }
                //continue;
            } else if(points[i].type =='B'){
                if(result>=0){
                    B1++;
                }else{
                    B0++;
                }
                //continue;
            } 
        }
        if(A0==count_A&&B1==count_B){
            //A全小于0，B全大于0
             cout<<"Yes"<<endl;
             continue;
        }
        if(A1==count_A&&B0==count_B){
            //A全大于0，B全小于0
             cout<<"Yes"<<endl;
             continue; 
        }         
        cout<<"No"<<endl;
    }
    return 0;
}