2010-2011 ACM-ICPC, NEERC, Moscow Subregional Contest Problem A. Alien Visit 计算几何

Problem A. Alien Visit

题目连接:

http://codeforces.com/gym/100714

Description

Witness: “First, I saw only one UFO. It was shining with cold-blue light. Closer to the center of the
object, the light was pink. It hung over the field, then began to blink and move intermittently round. The
UFO was quite big. The second UFO came several minutes after the first. It had the same size as the first
one. It seemed that there was some kind of contact between them — they began to blink alternately.”
Circles of scorched barley were found in the field. The circles were of the same radius, and their centers
were lying on a straight line.
You were hired to investigate the damage caused to the farms of Elcino-Borisovo place by the visit of
aliens. In order to do this you are to calculate the total area of scorched barley.

Input

The first line of the input contains two integers n and r denoting number of circles and the radius of the
circles, respectively (1 ≤ n ≤ 1 000, 1 ≤ r ≤ 100). The next line contains n space separated integers
a1, a2, . . . , an — the shifts of circles’ centers relative to some origin (0 ≤ ai ≤ 5 000). All shifts are
guaranteed to be distinct.

Output

Output the only real number — the total area covered by these circles. The relative error of your answer
must not exceed 10−6
.

Sample Input

1 1
0

Sample Output

3.1415926536

Hint

题意

给你n个在x轴的圆,半径都是r,问你总共的面积是多少

题解:

就总的面积减去两两相交的面积就好了,咩。

代码

#include<bits/stdc++.h>
using namespace std;
//两圆公共面积:
// 必须保证相交
const double PI = acos(-1.0);
struct POINT
{
 double x;
 double y;
 POINT(double a=0, double b=0) { x=a; y=b;} //constructor
};
double c2area(POINT A,double r1,POINT B,double r2){
    double rx1=A.x,ry1=A.y,rx2=B.x,ry2=B.y;
    double drma=sqrt((rx1-rx2)*(rx1-rx2)+(ry1-ry2)*(ry1-ry2));
    double a1=acos((r1*r1+drma*drma-r2*r2)/(2.0*r1*drma));
    double a2=acos((r2*r2+drma*drma-r1*r1)/(2.0*r2*drma));
    double res=(a1*r1*r1+a2*r2*r2-r1*drma*sin(a1));
    return res;
}
double dis(POINT A,POINT B){
    return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}
bool cmp(POINT A,POINT B){
    if(A.x==B.x)return A.y<B.y;
    return A.x<B.x;
}
POINT p[5005];
int main(){
    int n;double r;
    scanf("%d%lf",&n,&r);
    for(int i=1;i<=n;i++){
        scanf("%lf",&p[i].x);
        p[i].y=0;
    }
    sort(p+1,p+1+n,cmp);
    double ans = n*PI*r*r;
    for(int i=1;i<n;i++){
        if(2*r>dis(p[i],p[i+1])){
            ans-=c2area(p[i],r,p[i+1],r);
        }
    }
    printf("%.12f\n",ans);
}
posted @ 2016-08-08 19:41  qscqesze  阅读(319)  评论(0编辑  收藏  举报