poj 1385 Lifting the Stone 多边形求重心

题目链接

知识讲解

两种代码~

#include<stdio.h>
#include<iostream>
#include<cmath>
using namespace std;
const double eps=1e-7;
int cmp(double a){
     if(fabs(a)<eps)return 0;
     else return 1;
     }
struct node{
       double x,y;
       node(){}
       node(double a,double b):x(a),y(b){}
       friend node operator +(const node &a, const node &b){      
             return node( a.x+b.x, a.y+b.y);
       }
       friend node operator -(const node &a, const node &b){      
             return node( a.x-b.x, a.y-b.y);
       }
       friend node operator *(const node &a, const double &b  ){
             return node( a.x*b, a.y*b);      
       }
       friend node operator *(const double &b , const node &a ){
             return node( a.x*b, a.y*b);      
       }
       friend node operator /(const node a,const double &b){
             return node( a.x/b, a.y/b);       
       }
};
double fabs_det(node a,node b){   //求面积 
       return a.x*b.y-a.y*b.x;       
}

struct convex{
       node point[1200000];
       int n;
       double area(){
             point[n]=point[0];
             double sum=0;
             for(int i=0;i<n;i++){
                   sum+= (point[i].x*point[i+1].y-point[i].y*point[i+1].x);        
             }
             return sum/2; 
       }    
       node get_point(){
            node ans(0,0);
            //point[n]=point[0];
            for(int i=1;i<n-1;i++){
               ans=ans+fabs_det(point[i]-point[0],point[i+1]-point[0])*(point[i+1]+point[i]+point[0]);  //这里的叉积和面积都不要求绝对指
            }
            return ans/ (area()*6);            
       }
       
       
       
}shape;


int main()
{  
     int x;
     scanf("%d",&x);
     while(x--){
                
         scanf("%d",&shape.n);
         for (int i=0;i<shape.n;i++){
             scanf("%lf%lf",&shape.point[i].x,&shape.point[i].y);    
         }  
         //printf("%lf\n",shape.area());                              
         node ans=shape.get_point();
         printf("%.2lf %.2lf\n",cmp(ans.x)==0?0.00:ans.x,cmp(ans.y)==0?0.00:ans.y);                            
     }
    return 0;    
}

第二中方法

#include<stdio.h>
#include<iostream>
#include<cmath>
using namespace std;
const double eps=1e-7;
int cmp(double a){
     if(fabs(a)<eps)return 0;
     else return 1;
     }
struct node{
       double x,y;
       node(){}
       node(double a,double b):x(a),y(b){}
       friend node operator +(const node &a, const node &b){        //重载对点的运算
             return node( a.x+b.x, a.y+b.y);
       }
       friend node operator -(const node &a, const node &b){      
             return node( a.x-b.x, a.y-b.y);
       }
       friend node operator *(const node &a, const double &b  ){
             return node( a.x*b, a.y*b);      
       }
       friend node operator *(const double &b , const node &a ){
             return node( a.x*b, a.y*b);      
       }
       friend node operator /(const node &a,const double &b){
             return node( a.x/b, a.y/b);       
       }
};
double det(node a,node b){   //求面积 叉积
       return a.x*b.y-a.y*b.x;       
}

struct convex{
       node point[1200000];
       int n;
       double area(){
             point[n]=point[0];
             double sum=0;
             for(int i=0;i<n;i++){
                   sum+= (point[i].x*point[i+1].y-point[i].y*point[i+1].x);        
             }
             return sum/2; 
       }    
       node get_point(){
            node ans(0,0);
            point[n]=point[0];
            for(int i=0;i<n;i++){
               ans=ans+  (point[i]+point[i+1]) * det(point[i],point[i+1]) ;   //不要求绝对值
            }
            return ans/ (area()*6);            
       }
       
       
       
}shape;


int main()
{  
     int x;
     scanf("%d",&x);
     while(x--){
         scanf("%d",&shape.n);
         for (int i=0;i<shape.n;i++){
             scanf("%lf%lf",&shape.point[i].x,&shape.point[i].y);    
         }  
         //printf("%lf\n",shape.area());                              
         node ans=shape.get_point();
         printf("%.2lf %.2lf\n",cmp(ans.x)==0?0.00:ans.x,cmp(ans.y)==0?0.00:ans.y);                            
     }
    return 0;    
}