HDU 3264 Open-air shopping malls （計算幾何-圓相交面積）

傳送門：http://acm.hdu.edu.cn/showproblem.php?pid=3264

給你n個圓，座標和半徑，然後要在這n個圓的圓心畫一個大圓，大圓與這n個圓相交的面積必須大於等於每個圓面積的一半，問你建在那個圓心半徑最小，為多少。

枚舉這n個圓，求每個圓的最小半徑，通過二分半徑來求，然後取這n個的最小值即可，注意點精度就OK了。

AC代碼：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;#define si1(a) scanf("%d",&a)#define si2(a,b) scanf("%d%d",&a,&b)#define sd1(a) scanf("%lf",&a)#define sd2(a,b) scanf("%lf%lf",&a,&b)#define ss1(s)  scanf("%s",s)#define pi1(a)    printf("%d\n",a)#define pi2(a,b)  printf("%d %d\n",a,b)#define mset(a,b)   memset(a,b,sizeof(a))#define forb(i,a,b)   for(int i=a;i<b;i++)#define ford(i,a,b)   for(int i=a;i<=b;i++)typedef long long LL;const int N=33;const int INF=0x3f3f3f3f;const double PI=acos(-1.0);const double eps=1e-8;int n;struct xkn{    double x,y,r;    double area;}p[22],h;double dis(xkn a,xkn b){    return sqrt( (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) );}double fuck(xkn a,xkn b)  //求兩圓的相交面積函數。{    double d=dis(a,b);    if(d>=a.r+b.r) return 0;    double r=(a.r>b.r?b.r:a.r);    if( d<=fabs(a.r-b.r) ) return PI*r*r;    double A1=acos( (a.r*a.r+d*d-b.r*b.r)/2/a.r/d );    double A2=acos( (b.r*b.r+d*d-a.r*a.r)/2/b.r/d );    double res=A1*a.r*a.r + A2*b.r*b.r;    res-=sin(A1)*a.r*d;    return res;}bool xiaohao(xkn h){    for(int i=0;i<n;i++)    {        double jiao=fuck(p[i],h);        if(jiao<p[i].area/2)            return false;    }    return true;}int main(){//    freopen("input.txt","r",stdin);    int T;    si1(T);    while(T--)    {        si1(n);        for(int i=0;i<n;i++)        {            sd2(p[i].x,p[i].y);            sd1(p[i].r);            p[i].area=PI*p[i].r*p[i].r;        }        double Min=55555;        for(int i=0;i<n;i++)        {            double l=0,r=22222,m;            while((r-l)>eps)            {                m=(l+r)/2;                h=p[i]; h.r=m;                if(xiaohao(h))                    r=m;                else                    l=m;            }            if(Min>m)   Min=m;        }        printf("%.4f\n",Min);    }    return 0;}//92539932013-09-30 19:54:36Accepted32640MS320K2260 BG++XH_Reventon