HDU 1115 Lifting the Stone（求多边形重心）

HDU 1115

大意：给你个n，有n个点，然后给你n个点的坐标，求这n个点形成的多边形的重心的坐标。

struct point
{
    double x, y;
} P[1000010];

struct line
{
    point a, b;
} ;

double xmult(point p1, point p2, point p)
{
    return (p1.x-p.x)*(p2.y-p.y) - (p2.x-p.x)*(p1.y-p.y);
}

point intersection(line u,line v){
    point ret=u.a;
    double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))
            /((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
    ret.x+=(u.b.x-u.a.x)*t;
    ret.y+=(u.b.y-u.a.y)*t;
    return ret;
}

point barycenter(point a,point b,point c){
    line u,v;
    u.a.x=(a.x+b.x)/2;
    u.a.y=(a.y+b.y)/2;
    u.b=c;
    v.a.x=(a.x+c.x)/2;
    v.a.y=(a.y+c.y)/2;
    v.b=b;
    return intersection(u,v);
}

///多边形重心
point Barycenter(int n,point* p){
    point ret,t;
    double t1=0,t2;
    int i;
    ret.x=ret.y=0;
    for (i=1;i<n-1;i++)
        if (fabs(t2=xmult(p[0],p[i],p[i+1]))>eps){
            t=barycenter(p[0],p[i],p[i+1]);
            ret.x+=t.x*t2;
            ret.y+=t.y*t2;
            t1+=t2;
        }
    if (fabs(t1)>eps)
        ret.x/=t1,ret.y/=t1;
    return ret;
}

int T;
int n;

void Solve()
{
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; ++i)
        {
            scanf("%lf%lf", &P[i].x, &P[i].y);
        }
        point t = Barycenter(n, &P[0]);
        printf("%.2lf %.2lf
", t.x+eps, t.y+eps);
    }
}