【TOJ 3005】Triangle（判断点是否在三角形内+卡精度）

描述

Given the coordinates of the vertices of a triangle，And a point. You just need to judge whether the point is in the Triangle.

输入

The input contains several test cases. For each test case, only line contains eight integer numbers , describing the coordinates of the triangle and the point. All the integer is in range of [-100 , 100].
The end of the input is indicated by a line containing eight zeros separated by spaces.

输出

For each test case , if the point is inside of the triangle ,please output the string ”YES”, else output the string “NO”. You just need to follow the following examples.

样例输入

0 0 4 0 0 4 3 1
0 0 4 0 0 4 1 2
0 0 4 0 0 4 -1 -1
0 0 0 0 0 0 0 0

样例输出

NO
YES
NO

思路：

通过判断3个小三角形面积是否等于大三角形面积即可，这道题卡精度，要控制误差在1e-9。

#include<bits/stdc++.h>
using namespace std;
struct node{
    int x,y;
}a,b,c,p;
double edge(int x1,int y1,int x2,int y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double area(int x1,int y1,int x2,int y2,int x3,int y3)
{
    double A=edge(x1,y1,x2,y2);
    double B=edge(x1,y1,x3,y3);
    double C=edge(x3,y3,x2,y2);
    double L=(A+B+C)*1.0/;
    return fabs(sqrt(L*(L-A)*(L-B)*(L-C)));
}
bool check(double a,double b,double c,double p)
{
    if(a==||b==||c==)
        return false;
    if(fabs(a+b+c-p)<=1e-)
        return true;
    return false;
}
int main()
{
    while(cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y>>p.x>>p.y,a.x||a.y||b.x||b.y||c.x||c.y||p.x||p.y)
    {
        double s=area(a.x,a.y,b.x,b.y,c.x,c.y);
        double abp=area(a.x,a.y,b.x,b.y,p.x,p.y);
        double acp=area(a.x,a.y,c.x,c.y,p.x,p.y);
        double bcp=area(c.x,c.y,b.x,b.y,p.x,p.y);

        if(check(abp,acp,bcp,s))
            printf("YES\n");
        else printf("NO\n");
    }
    return ;
}