zoukankan      html  css  js  c++  java
  • UVA 10002 Center of Masses

    题目链接:http://acm.uva.es/local/online_judge/search_uva.html

    Problem:Find out the center of masses of a convex polygon.

    Input:A series of convex polygons, defined as a number n ($n le 100$) stating the number of points of the polygon, followed by n different pairs of integers (in no particular order), denoting the x and y coordinates of each point. The input is finished by a fake ``polygon" with m (m < 3) points, which should not be processed.

    Output:For each polygon, a single line with the coordinates x and y of the center of masses of that polygon, rounded to three decimal digits.

    解法:求多边形的重心。先求出其凸包,然后求重心。

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cstdlib>
     5 #include<cmath>
     6 #include<algorithm>
     7 #define inf 0x7fffffff
     8 #define exp 1e-10
     9 #define PI 3.141592654
    10 using namespace std;
    11 const int maxn=111;
    12 struct Point
    13 {
    14     double x,y;
    15     Point (double x=0,double y=0):x(x),y(y){}
    16     bool friend operator < (Point a,Point b)
    17     {
    18         if (a.x!=b.x) return a.x<b.x;
    19         return a.y<b.y;
    20     }
    21 }an[maxn],bn[maxn];
    22 typedef Point Vector;
    23 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); }
    24 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); }
    25 Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); }
    26 int dcmp(double x)
    27 {
    28     if (fabs(x)<exp) return 0;
    29     return x<0 ? -1 : 1;
    30 }
    31 double cross(Vector A,Vector B)
    32 {
    33     return A.x*B.y-B.x*A.y;
    34 }
    35 Point PolyGravity(Point *p,int n)
    36 {
    37     Point tmp,g=Point(0,0);
    38     double sumArea=0,area;
    39     for (int i=2 ;i<n ;i++)
    40     {
    41         area=cross(p[i-1]-p[0],p[i]-p[0]);
    42         sumArea += area;
    43         tmp.x=p[0].x+p[i-1].x+p[i].x;
    44         tmp.y=p[0].y+p[i-1].y+p[i].y;
    45         g.x += tmp.x*area;
    46         g.y += tmp.y*area;
    47     }
    48     g.x /= (sumArea*3.0);
    49     g.y /= (sumArea*3.0);
    50     return g;
    51 }
    52 int ConvexHull(Point *p,int n,Point *ch)
    53 {
    54     sort(p,p+n);
    55     int m=0;
    56     for (int i=0 ;i<n ;i++)
    57     {
    58         while (m>1 && dcmp(cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<0) m--;
    59         ch[m++]=p[i];
    60     }
    61     int k=m;
    62     for (int i=n-2 ;i>=0 ;i--)
    63     {
    64         while (m>k && dcmp(cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<0) m--;
    65         ch[m++]=p[i];
    66     }
    67     if (n>1) m--;
    68     return m;
    69 }
    70 int main()
    71 {
    72     int n;
    73     while (scanf("%d",&n)!=EOF)
    74     {
    75         if (n<3) break;
    76         for (int i=0 ;i<n ;i++)
    77         {
    78             scanf("%lf%lf",&an[i].x,&an[i].y);
    79         }
    80         int k=ConvexHull(an,n,bn);
    81         //cout<<"DEBUG
    ";
    82         //for (int i=0 ;i<n ;i++) cout<<an[i].x<<" "<<an[i].y<<endl;
    83         Point ans=PolyGravity(bn,k);
    84         printf("%.3lf %.3lf
    ",ans.x,ans.y);
    85     }
    86     return 0;
    87 }
  • 相关阅读:
    Java中,&&与&,||与|的区别
    Hibernate中的merge方法 以及对象的几中状态
    你希望函数的某些参数强制使用关键字参数传递:
    7.1 可接受任意数量参数的函数:
    perl urlencode
    python UnicodeDecodeError: 'utf8' codec can't decode byte 0xd6 in position 15: invalid continuation
    python 使用__slots__
    python 面向对象编程
    Python flask post接口
    python flask get传参
  • 原文地址:https://www.cnblogs.com/huangxf/p/3928397.html
Copyright © 2011-2022 走看看