2013省赛集训5 - Triangle

最新推荐文章于 2019-09-16 11:43:03 发布

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本文介绍了一种判断二维平面上一点是否位于给定三角形内部的有效算法，并提供了完整的C++实现代码。该算法利用三角形有向面积的概念来确定点的位置。

Triangle

Time Limit 1000ms

Memory Limit 65536K

description

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

input

　　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.

output

　　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.

sample_input

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

sample_output

NO
YES
NO

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>

using namespace std;

int area(int x1,int y1,int x2,int y2,int x3,int y3)
{
    return x1*y2+x2*y3+x3*y1-x2*y1-x3*y2-x1*y3;
}//三角形有向面积的2倍，逆时针正，顺时针负

int main()
{
    int x1,y1,x2,y2,x3,y3,x,y,s1,s2,s3,s;
    while(scanf("%d%d%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3,&x,&y)!=EOF)
    {
        if(x1==0&&y1==0&&x2==0&&y2==0&&x3==0&&y3==0&&x==0&&y==0)
            break;
        x1+=100;y1+=100;x2+=100;y2+=100;x3+=100;y3+=100;x+=100;y+=100;
        s1=abs(area(x1,y1,x2,y2,x,y));
        s2=abs(area(x1,y1,x3,y3,x,y));
        s3=abs(area(x2,y2,x3,y3,x,y));
        s=abs(area(x1,y1,x2,y2,x3,y3));
        if((s1+s2+s3)==s&&s1&&s2&&s3)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}