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 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.
For each transmitter, the output contains a single line with the maximum number of points that can be contained in some semicircle.
Example 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
Example output:
3
4
4
题目大意：给一个半圆的圆心和半径，然后给出num个点(x,y),其中半圆可以绕轴心旋转，问最多能有几个点被覆盖，当输入圆的半径为负数时输入结束。
解题思路：直接暴力，每次以圆心和一个点的连线为直径通过差乘>=0为半径的一侧(包含直径上),然后通过判断满足上面的点到圆心的距离和半径的距离判断是否在圆内。
相关知识：差乘与点乘http://blog.csdn.net/zxj1988/article/details/6260576
1 #include<iostream> 2 #include<string> 3 using namespace std; 4 double Yuan_x,Yuan_y,Yuan_r; 5 int num,Maxnum; 6 double x[155],y[155]; 7 bool read(){ 8 cin>>Yuan_x>>Yuan_y>>Yuan_r; 9 if(Yuan_r<0)return 0; 10 cin>>num; 11 for(int i=0;i<num;i++)cin>>x[i]>>y[i]; 12 return 1; 13 } 14 void solve(){ 15 Maxnum=-1; 16 int temp; 17 for(int i=0;i<num;i++){ 18 temp=0; 19 for(int j=0;j<num;j++){ 20 int v_x1=x[i]-Yuan_x , v_y1=y[i]-Yuan_y , 21 v_x2=x[j]-Yuan_x , v_y2=y[j]-Yuan_y ; 22 if(v_x1*v_y2-v_x2*v_y1<0)continue; 23 else temp+=(v_x2*v_x2+v_y2*v_y2<=Yuan_r*Yuan_r); 24 } 25 if(temp>Maxnum)Maxnum=temp; 26 } 27 }//直接暴力枚举 28 int main(){ 29 while(read()){ 30 solve(); 31 cout<<Maxnum<<'\n'; 32 }return 0; 33 }
