POJ Transmitters（计算几何 极角排序啊）

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

Output

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。

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;
}

POJ Transmitters（计算几何 极角排序啊）

原文：http://blog.csdn.net/u012860063/article/details/45876319