J - Transmitters解题报告（熊禾强）

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探讨了无线网络中如何通过调整屏蔽型半圆发射器的方向来最大化覆盖特定网格上的数据点数量。利用几何方法和叉积算法确定最佳发射角度。

J - Transmitters

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 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

题目大意：给出圆心坐标和半径，并给出几个点，求出半圆放在哪个角度是包含的给出的点最多。

解题思路:叉积算法，设矢量P=（x1,y1），Q=（x2,y2），若P*Q>0，则P在Q的顺时针方向，

若P*Q<0,则P在Q逆时针方向。先判断点是否在圆内，再用叉积。

若P*Q=0，则P和Q共线

代码：

#include<stdio.h> 
#include<math.h> 
const double inf=1e-6; 
double x[160],y[160]; 
double dis(int xa,int ya,int xb,int yb) 
{ 
  return sqrt(double((xa-xb)*(xa-xb)+(ya-yb)*(ya-yb))); //判断点是否在圆内
} 
int side(int i,int j,int xx,int yy) 
{ 
 
if(	(xx - x[i]) * (yy - y[j]) - (xx - x[j]) * (yy - y[i]) >= 0)//叉积判断 
return 1; 
else 
return 0;
} 
int main() 
{ 
  double r;
  int ox,oy,xx,yy,n,i; 
  while(scanf("%d%d%lf",&ox,&oy,&r)!=EOF && r>0) 
  { 
	    scanf("%d",&n); 
	       int j=0; 
	        for( i=0;i<n;i++) 
	        { 
	            scanf("%d%d",&xx,&yy); 
	          if(r-dis(xx,yy,ox,oy)>-inf)//判断点是否在圆内 
	            { 
	                x[j]=xx; 
	                y[j]=yy; 
	                ++j; 
	            } 
	        } 
	        n=j; 
	        int maxx=0; 
	        for( i=0;i<n;i++) //判断在哪个角度，包含的点最多
	        { 
	            int k=0; 
	            for(int j=0;j<n;j++) 
	                if(side(i,j,ox,oy)) 
	                    ++k; 
	            if(maxx<k)maxx=k; 
	        } 
	        printf("%d\n",maxx); 
	    }
		return 0;
	}