UVA 2290 Transmitters

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=291

cross product: for 2 vectors a and b, if a^b > 0, a is in the clockwise direction of b. else if a^b < 0, a is in the anti-clockwise direction of vector b. Otherwise, a^b==0, vector a and b are with the same direction or opposite direction.这道题用cross product计算。每一个点与transmitter的中心形成一个vector，对于每一个vector，循环剩余的vector，所有vector在同一方向的可以看做被覆盖的点。代码如下：

 //============================================================================
 // Name        : test.cpp
 // Author      :
 // Version     :
 // Copyright   : Your copyright notice
 // Description : Hello World in C++, Ansi-style
 //============================================================================

 #include <iostream>
 #include <math.h>
 #include <stdio.h>
 #include <cstdio>
 #include <algorithm>
 #include <string.h>
 #include <string>
 #include <sstream>
 #include <cstring>
 #include <queue>
 #include <vector>
 #include <functional>
 #include <cmath>
 #include <set>
 #define SCF(a) scanf("%d", &a)
 #define IN(a) cin>>a
 #define FOR(i, a, b) for(int i=a;i<b;i++)
 #define Infinity 999999999
 #define PI 3.14159265358979323846
 typedef long long Int;
 using namespace std;

 struct point {
     int x, y;
 };

 double dis(point a, point b)
 {
     return pow(a.x - b.x, ) + pow(a.y - b.y, );
 }

 double length(point a)
 {
     return sqrt(pow(a.x, ) + pow(a.y, ));
 }

 double dotProduct(point a, point b)
 {
     return a.x*b.x + a.y*b.y;
 }

 double crossProduct(point a, point b)
 {
     return a.x*b.y - b.x*a.y;
 }

 double cosAngle(point a, point b)
 {
     double an = dotProduct(a, b) / (length(a)*length(b));
     return an;
 }

 int main()
 {
     point radar;
     double r;
     int N;
     point points[];
     double angle[];
     while (scanf("%d %d %lf", &radar.x, &radar.y, &r) != EOF)
     {
         if (r < )
             break;
         SCF(N);
         int index = ;
         point cp;
         FOR(i, , N)
         {
             SCF(cp.x);
             SCF(cp.y);
             if (dis(radar, cp) <= pow(r, ))
             {
                 points[index].x = cp.x - radar.x;
                 points[index++].y = cp.y - radar.y;
             }
         }

         int maxAns = ;
         FOR(i, , index)
         {
             int cn = ;
             FOR(j, , index)
             {
                 if (i != j)
                 {
                     double rela = crossProduct(points[i], points[j]);
                     if (rela >= )
                         cn++;
                 }
             }
             maxAns = max(maxAns, cn);
         }
         printf("%d\n", maxAns);
     }
     return ;
 }