ZOJ 3147 Dezider's Coverup【最小圆覆盖】【模板题】

Dezider's Coverup

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Zlatica to his house for dinner. The trouble is, he tripped with the sauce in his hands (thankfully this was in the morning and Zlatica did not see it) and many bright red droplets fell on his white carpet. Now Dezider is starting to panic because he wants to have a perfectly clean house to impress Zlatica. Fortunately, there is a store nearby which sells all sizes of (expensive) circular rugs. Dezider does not have much money so he wants to buy the smallest rug with an integer radius which will cover all the droplets. You need to write a program to help him determine the right size to buy.

Input

Each test cases starts with one positive integer n, the number of droplets on the carpet. Each of the following n lines contains two floating-point numbers, separated by white space, which represent the x- and y-coordinate of the corresponding droplet.

Output

One line for each test case which consists of a single integer, the radius of the smallest circular rug needed to cover all the droplets.

Sample Input

4
1 -0.5
1.5 0
0.5 0.5
1 0.5

Sample Output

1

because a circular rug of radius 1 and center for example at 1,0 will cover the four droplets described by the input.

#include <stdio.h>
#include <math.h>
const int maxn=1005;
const double eps=1e-9;
struct point
{
    double x, y;
    point operator-(point & tp)
    {
        point rp;
        rp.x=x-tp.x;
        rp.y=y-tp.y;
        return rp;
    }
}p[maxn];
struct circle
{
    double r;
    point centre;
}c;
struct tripoint
{
    point t[3];
};
double dis(point a, point b)
{
    a=a-b;
    return sqrt(a.x * a.x + a.y * a.y);
}
double triangleArea(tripoint t)
{
    point p1, p2;
    p1=t.t[1]-t.t[0];
    p2=t.t[2]-t.t[0];
    return fabs(p1.x*p2.y-p1.y*p2.x)/2;
}
circle circumcircleOfTriangle(tripoint t)
{    //三角形的外接圆
    circle tmp;
    double a, b, c, c1, c2;
    double xA, yA, xB, yB, xC, yC;
    a=dis(t.t[0], t.t[1]);
    b=dis(t.t[1], t.t[2]);
    c=dis(t.t[2], t.t[0]);    //根据S = a * b * c / R / 4;求半径R
    tmp.r=a*b*c/triangleArea(t)/4;
    xA=t.t[0].x;
    yA=t.t[0].y;
    xB=t.t[1].x;
    yB=t.t[1].y;
    xC=t.t[2].x;
    yC=t.t[2].y;
    c1=(xA*xA+yA*yA-xB*xB-yB*yB)/2;
    c2=(xA*xA+yA*yA-xC*xC-yC*yC)/2;
    tmp.centre.x=(c1*(yA-yC)-c2*(yA-yB))/((xA-xB)*(yA-yC)-(xA-xC)*(yA-yB));
    tmp.centre.y=(c1*(xA-xC)-c2*(xA-xB))/((yA-yB)*(xA-xC)-(yA-yC)*(xA-xB));
    return tmp;
}
circle MinCircle2(int tce, tripoint ce)
{
    circle tmp;
    if(tce==0)
        tmp.r=-2;
    else if(tce==1)
    {
        tmp.centre=ce.t[0];
        tmp.r=0;
    }
    else if(tce==2)
    {
        tmp.r=dis(ce.t[0], ce.t[1])/2;
        tmp.centre.x=(ce.t[0].x+ce.t[1].x)/2;
        tmp.centre.y=(ce.t[0].y+ce.t[1].y)/2;
    }
    else if(tce==3)
        tmp=circumcircleOfTriangle(ce);
    return tmp;
}
void MinCircle(int t, int tce, tripoint ce)
{
    int i, j;
    point tmp;
    c=MinCircle2(tce, ce);
    if(tce==3) return;
    for(i=1; i<=t; i++)
    {
        if(dis(p[i], c.centre)>c.r)
        {
            ce.t[tce]=p[i];
            MinCircle(i-1, tce+1, ce);
            tmp=p[i];
            for(j=i; j>=2; j--)
                p[j]=p[j-1];
            p[1]=tmp;
        }
    }
}
int main()
{
    int n,i;
    while(~scanf("%d", &n) && n)
    {
        for(i=1; i<=n; i++)
            scanf("%lf%lf", &p[i].x, &p[i].y);
        tripoint ce;
        MinCircle(n, 0, ce);
        int x=(int)c.r;
        if(c.r-x<eps)
        {
            printf("%d\n",x);
        }
        else printf("%d\n",x+1);
    }
    return 0;
}