题目链接：http://poj.org/problem?id=1106
Description
In a wireless network with multiple transmitters sending on the same frequencies, it is often a requirement that signals don‘t overlap, or at least that they don‘t conflict. One way of accomplishing this is to restrict a transmitter‘s coverage area. This problem
uses a shielded transmitter that only broadcasts in a semicircle.
A transmitter T is located somewhere on a 1,000 square meter grid. It broadcasts in a semicircular area of radius r. The transmitter may be rotated any amount, but not moved. Given N points anywhere on the grid, compute the maximum number of points that can
be simultaneously reached by the transmitter‘s signal. Figure 1 shows the same data points with two different transmitter rotations.
All input coordinates are integers (0-1000). The radius is a positive real number greater than 0. Points on the boundary of a semicircle are considered within that semicircle. There are 1-150 unique points to examine per transmitter. No points are at the same
location as the transmitter.
Input
Input consists of information for one or more independent transmitter problems. Each problem begins with one line containing the (x,y) coordinates of the transmitter followed by the broadcast radius, r. The next line contains the number of points N on the grid,
followed by N sets of (x,y) coordinates, one set per line. The end of the input is signalled by a line with a negative radius; the (x,y) values will be present but indeterminate. Figures 1 and 2 represent the data in the first two example data sets below,
though they are on different scales. Figures 1a and 2 show transmitter rotations that result in maximal coverage.
Output
For each transmitter, the output contains a single line with the maximum number of points that can be contained in some semicircle.
Sample Input
25 25 3.5 7 25 28 23 27 27 27 24 23 26 23 24 29 26 29 350 200 2.0 5 350 202 350 199 350 198 348 200 352 200 995 995 10.0 4 1000 1000 999 998 990 992 1000 999 100 100 -2.5
Sample Output
3 4 4
Source
题意：
给定一些点，和一个圆心坐标和半径，求一个半圆能最多能圈下多少点，半圆可绕着圆心任意旋转。
PS:
只需要考虑到圆心距离小于或者等于半径的那些点，
把符合条件的点全部存入一个数组tt[]，
然后二重循环枚举每一个点与圆心所连的直线的的左侧有多少个点（叉积），
记录最大值即可。
叉积丶点积：http://blog.csdn.net/y990041769/article/details/38258761
代码如下：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
struct node
{
double x, y;
} p[100017],tt[100017];
double cross(node A,node B,node C)//叉积
{
return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);
}
double dis(node A,node B)//距离
{
return (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y);
}
int main()
{
double r;
node c;
while(~scanf("%lf%lf%lf",&c.x,&c.y,&r))
{
if(r < 0)
{
break;
}
int n;
scanf("%d",&n);
int k = 0;
for(int i = 0; i < n; i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
double d = dis(c,p[i]);
if(d <= r*r)
{
tt[k].x = p[i].x;
tt[k].y = p[i].y;
k++;
}
}
int ansmax = 0;
for(int i = 0; i < k; i++)
{
int cont = 1;
for(int j = 0; j < k; j++)
{
if(i!= j && cross(c,tt[i],tt[j])>=0)
{
cont++;
}
}
if(cont > ansmax)
{
ansmax = cont;
}
}
printf("%d\n",ansmax);
}
return 0;
}