Time Limit: 1000ms Memory Limit: 512MB Description A light comes from point s, passed from point e, and finally go to infinite far. There is a light given by its s and e. You’re given a point, and your task is to judge if the point is on this light. Input First line contains an integer T (1 ≤ T ≤ 10v) represents the number of test cases. For each test case: The first line contains four integers sx, sy, ex, ey of the light. The second line contains two integers px, py of the point. (1 ≤ sx, sy, ex, ey, px, py ≤ 1000 and (sx, sy) ! = (ex, ey)). Output For each test case, if the point is on the light, output “YES” in a single line; otherwise, output “NO” in a single line. Sample Input 4 2 2 3 3 2 2 2 2 3 3 1 1 7 7 11 10 855 643 7 7 7 8 7 9 Output YES NO YES YES
题目大意： 从 s s s点发出的光，经过 e e e点。现给出 s s s点和 e e e点的坐标，再给一个 p p p点，让你判断p点是否在光线上。 分析： 很简单的几何问题。 由题可知，光线是一个射线。判断 p p p点是否在光线上时，注意判断方向。 思路： 采用向量法求解：先求出 p s ⃗ \vec{ps} ps 和 e s ⃗ \vec{es} es 的坐标，由于两个向量都包含光源 s s s点，故直接判断它们是否同向即可。 代码如下：
#include <iostream> using namespace std; //点的结构体 struct point { int x, y; //点的横纵坐标 }; int t; point s, e, p; //判断点p是否在射线s->e上 int main() { scanf("%d", &t); while(t--) { scanf("%d%d%d%d", &s.x, &s.y, &e.x, &e.y); scanf("%d%d", &p.x, &p.y); //光线是一条射线，由于射线只朝着一个方向传播，我用向量法判断方向是否一致 //1. 求向量PS和ES的坐标 int psx = p.x - s.x, psy = p.y - s.y; //向量PS = (psx, psy) int esx = e.x - s.x, esy = e.y - s.y; //向量ES = (esx, esy) //2. 如果PS和ES不平行或平行但反向，则必不在光线上，反之则在光线上。 //x1*y2 - x2*y1 ！= 0: 代表两个向量不平行， x1 * x2 < 0: 代表两个向量反向; if(psx * esy - esx * psy || psx * esx < 0) printf("NO\n"); else printf("YES\n"); } return 0; }