Coverage 圆与直线的交点

最新推荐文章于 2020-12-20 14:28:11 发布

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本文介绍了一种计算手机用户在特定路径上获得信号覆盖率的方法。通过解析路径与多个信号塔的位置和传输半径，利用韦达定理求解路径与信号塔覆盖区域的交集，最终得出信号覆盖的百分比。

Coverage

Time Limit: 1000ms

Memory Limit: 32768KB

64-bit integer IO format: %I64d Java class name: Main

Submit Status PID: 7536

A cell phone user is travelling along a line segment with end points having integer coordinates. In order for the user to have cell phone coverage, it must be within the transmission radius of some transmission tower. As the user travels along the path, cell phone coverage may be gained (or lost) as the user moves inside the radius of some tower (or outside of the radii of all towers). Given the location of up to 100 towers and their transmission radii, you are to compute the percentage of cell phone coverage the user has
along the specified path. The (x,y) coordinates are integers between -100 and 100, inclusive, and the tower radii are integers between 1 and 100, inclusive.

Input

Your program will be given a sequence of configurations, one per line, of the form: N C0X C0Y C1X C1Y T1X T1Y T1R T2X T2Y T2R ... Here, N is the number of towers, (C0X,C0Y) is the start of path of the cell phone user,
(C1X,C1Y) is the end of the path, (TkX,TkY) is the position of the kth tower, and TkR is its transmission radius. The start and end points of the paths are distinct. The last problem is terminated by the line 0

Output

For each configuration, output one line containing the percentage of coverage the cell phone has, rounded to two decimal places.

Sample Input

3 0 0 100 0 0 0 10 5 0 10 15 0 10
1 0 0 100 0 40 10 50
0

Sample Output

25.00
88.99

分析：给出一条路径再给出n个塔及每个塔发出的信号波长，求被信号覆盖的路所占总路程的百分比；

韦达定理求线与圆的交点；

PS:当直线斜率不存在的时候将直角坐标系翻转90度即可以在输入的过程中将点的横纵坐标对调即可保证斜率不存在的情况下也能表示出斜率

<span style="font-size:18px;">#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
struct gg
{
    double x,y,r,lx,rx;
} a[105];
bool cmp(gg a,gg b)
{
    return a.lx<b.lx;
}
int main()
{
    int n;
    while (scanf("%d",&n))
    {
        if (n==0)break;
        double sx,sy,ex,ey,flag=0,temp;
        scanf("%lf%lf%lf%lf",&sx,&sy,&ex,&ey);
        if (sx==ex)
        {
            temp=sx;
            sx=sy;
            sy=temp;
            temp=ex;
            ex=ey;
            ey=temp;
            flag=1;
        }
        double k,c;
        k=(ey-sy)/(ex-sx);
        c=ey-k*ex;
        for (int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf",&a[i].x,&a[i].y,&a[i].r);
            if(flag)
            {
                temp=a[i].x;
                a[i].x=a[i].y;
                a[i].y=temp;
            }
            double A,B,C,T;
            A=1+k*k;
            B=-2*a[i].x+2*k*c-2*a[i].y*k;
            C=c*c+a[i].y*a[i].y+a[i].x*a[i].x-2*a[i].y*c-a[i].r*a[i].r;
            T=B*B-4*A*C;
            if(T>0)
            {
                a[i].lx=(-B-sqrt(T))/(2*A);
                a[i].rx=(-B+sqrt(T))/(2*A);
                if(a[i].lx>max(sx,ex)||a[i].rx<min(sx,ex))//不在范围
                a[i].lx=a[i].rx=-111;
                else
                {
                    if(a[i].lx<min(sx,ex))a[i].lx=min(sx,ex);
                    if(a[i].rx>max(sx,ex))a[i].rx=max(sx,ex);
                }
            }
            else
                a[i].lx=a[i].rx=-111;

        }
        sort(a,a+n,cmp);
        double sum=0,l=a[0].lx,r=a[0].rx;
        for (int i=0; i<n; i++)
        {
            if(a[i].lx<r)
            {
                r=r>a[i].rx?r:a[i].rx;
            }
            else
            {
                sum+=r-l;
                l=a[i].lx;
                r=a[i].rx;
            }
            if (i==n-1)
                sum+=r-l;
        }
        printf("%.2lf\n",sum/((ex-sx)>0?(ex-sx):(sx-ex))*100);
    }
    return 0;
}</span>