codeforces #339 div 2 C. Peter and Snow Blower

Posted by 111qqz on Wednesday, February 17, 2016

TOC

http://codeforces.com/contest/614/problem/C

题意:给一个多边形和多边形外一定点,多边形绕定点旋转,问多边形扫过的面积。
思路:简单计算几何,找到多边形距离定点的最大和最小距离R和r,答案就是(R^2-R^2)*PI
需要注意的是:最大距离一定是从某点上取得,但是最小距离可能不在顶点上,而在某条边上。

/* ***********************************************
Author :111qqz
Created Time :2016年02月16日 星期二 16时58分59秒
File Name :code/cf/problem/614C.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N = 1E5+7;
const double PI = acos(-1.0);

int dblcmp(double d)
{
    return d<-eps?-1:d>eps;
}
struct point
{
    double x,y;
    point (){}
    point (double _x,double _y):
    x(_x),y(_y){};
    void input()
    {
    cin>>x>>y;
    }
    point sub(point p)
    {
    return point (x-p.x,y-p.y);
    }
    double dot(point p)
    {
    return x*p.x+y*p.y;
    }
    double distance2(point p)
    {
    return (x-p.x)*(x-p.x)+(y-p.y)*(y-p.y);
    }
    double distance(point p)
    {
    return hypot(x-p.x,y-p.y);
    }
    double det(point p)
    {
    return x*p.y-y*p.x;
    }
}p[N],q;
double sqr(double x)
{
    return x*x;
}
struct line
{
    point a,b;
    line(){}
    line (point _a,point _b)
    {
    a = _a;
    b = _b;
    }
    double length()
    {
    return a.distance(b);
    }

    double dispointtoline2(point p)
    {
    double res = sqr(fabs(p.sub(a).det(b.sub(a)))/length());
//	cout<<"res:"<<res<<endl;
    return res;
    }
    double dispointtoseg2(point p)
    {
    if (dblcmp(p.sub(b).dot(a.sub(b)))<0||dblcmp(p.sub(a).dot(b.sub(a)))<0)
        return min(p.distance2(a),p.distance2(b));
    
    return dispointtoline2(p);
    }
}li[N];

int n;
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif

    cin>>n;
    q.input();
    double MIN = 99999999999999999;
    double MAX = -1;
    for ( int i = 1 ; i <= n ; i++)
    {
        p[i].input();
        double d= p[i].distance2(q);
        MAX = max(MAX,d);
    }

    for ( int i = 1 ; i <= n-1 ; i++)
    {
        li[i] = line(p[i],p[i+1]);
    }
    li[n]=line(p[n],p[1]);
    for ( int i = 1 ; i <= n ; i++)
    {
        double d = li[i].dispointtoseg2(q);
//	    cout<<li[i].a.x<<" "<<li[i].a.y<<"   "<<li[i].b.x<<" "<<li[i].b.y<<endl; 
    //    cout<<"d:"<<d<<endl;
        MIN = min(MIN,d);
    }
//	cout<<"MAX:"<<MAX<<"  MIN:"<<MIN<<endl;
    double ans = (MAX-MIN)*PI;
    printf("%.18f\n",ans);


  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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