http://acm.hdu.edu.cn/showproblem.php?pid=1115
题意:求凸多边形的重心
思路:明明是SB题,做法就是三角剖分每个三角形的重心乘上三角形面积的矢量和,最后除以总面积,我愣是WA了3天,最后才知道要把所有的除数放到最后除才不会有精度误差,舞草
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #define dou double 7 struct Point{ 8 dou x,y; 9 Point(){} 10 Point(dou x0,dou y0):x(x0),y(y0){} 11 }p[5000005]; 12 int n; 13 int read(){ 14 int t=0,f=1;char ch=getchar(); 15 while (ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();} 16 while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();} 17 return t*f; 18 } 19 Point operator +(Point p1,Point p2){ 20 return Point(p1.x+p2.x,p1.y+p2.y); 21 } 22 dou operator *(Point p1,Point p2){ 23 return p1.x*p2.y-p1.y*p2.x; 24 } 25 Point operator *(Point p1,dou x){ 26 return Point(p1.x*x,p1.y*x); 27 } 28 Point operator /(Point p1,dou x){ 29 return Point(p1.x/x,p1.y/x); 30 } 31 Point operator -(Point p1,Point p2){ 32 return Point(p1.x-p2.x,p1.y-p2.y); 33 } 34 dou area(Point p1,Point p2,Point p3){ 35 return (p2-p1)*(p3-p1); 36 } 37 int main(){ 38 int T=read(); 39 while (T--){ 40 n=read(); 41 dou S=0;Point ans(0,0); 42 for (int i=1;i<=n;i++) 43 scanf("%lf%lf",&p[i].x,&p[i].y); 44 p[n+1]=p[1]; 45 for (int i=2;i<n;i++){ 46 S+=((p[i]-p[1])*(p[i+1]-p[1])); 47 ans=ans+((p[i]+p[i+1]+p[1]))*((p[i]-p[1])*(p[i+1]-p[1])); 48 } 49 ans=ans/(3.0*S); 50 double Ansx,Ansy; 51 Ansx=(double)(ans.x); 52 Ansy=(double)(ans.y); 53 printf("%.2f %.2f ",Ansx,Ansy); 54 } 55 }