uva10012 - How Big Is It?

想到可能存在两个半径很大的圆相切，中间夹着很多小的圆（这样单纯的全排列只判断相邻两圆的话就可能出现圆圆相交的情况），这样就需要加为每个已经判断过的圆加个坐标，对于正在判断的圆，与前面所有判断是否相交，若相交以这两个圆为相邻圆，求当前圆坐标。

#include <iostream>#include <cstring>#include<cstdio>#include<cstdlib>#include <string>#include<algorithm>#include<cmath>//#define MARK -2147483647using namespace std;double cir[10];int pai[10]={0,1,2,3,4,5,6,7,8,9}; int m;double ans=0;struct yuanxin{double x,y;};double qiulong( ){   yuanxin yuan[10];    double ll =0;    int i;    yuan[0].x=0.0;yuan[0].y=cir[pai[0]];    for(i=0;i<m-1;++i)    {   double llshi=ll;        //cout<<cir[pai[i]]<<cir[pai[i+1]]<<endl;        llshi=llshi+sqrt((cir[pai[i]]+cir[pai[i+1]])*(cir[pai[i]]+cir[pai[i+1]])-(cir[pai[i]]-cir[pai[i+1]])*(cir[pai[i]]-cir[pai[i+1]]));        //cout<<i<<"e"<<llshi<<endl;        yuan[i+1].x=llshi;yuan[i+1].y=cir[pai[i+1]];        for(int ii=i-1;ii>=0;--ii)        {            double bil=sqrt((yuan[ii].x-yuan[i+1].x)*(yuan[ii].x-yuan[i+1].x)+(yuan[ii].y-yuan[i+1].y)*(yuan[ii].y-yuan[i+1].y));            if(bil-(cir[pai[ii]]+cir[pai[i+1]])<-(1e-6))            {ll=yuan[ii].x+sqrt((cir[pai[ii]]+cir[pai[i+1]])*(cir[pai[ii]]+cir[pai[i+1]])-(cir[pai[ii]]-cir[pai[i+1]])*(cir[pai[ii]]-cir[pai[i+1]]));llshi=ll;yuan[i+1].x=llshi;   }        }//cout<<i<<"e"<<llshi<<endl;        ll=llshi;    }    ll=ll+cir[pai[m-1]];    for(int ii=m-2;ii>=0;--ii)    {        if(ll-(yuan[ii].x+cir[pai[ii]])<-(1e-6))ll=yuan[ii].x+cir[pai[ii]];    }    double bii=-cir[pai[0]];     for(int ii=m-1;ii>=0;--ii)    {        //if(ll-(yuan[ii].x+cir[ii])<-(1e-6))ll=yuan[ii].x+cir[ii];        if(bii-(yuan[ii].x-cir[pai[ii]])>1e-6) bii=yuan[ii].x-cir[pai[ii]];    }    return ll-bii;}int main(){// freopen("in.txt","r",stdin);    int n;scanf("%d",&n);    while(n--)    {        scanf("%d",&m);        int i;ans=0;        for(i=0;i<m;++i)        {            scanf("%lf",&cir[i]);            ans+=2*cir[i];        }        do        {            double ll=qiulong();            if((ll-ans)<-(1e-6)) ans=ll;        }while(next_permutation(pai,pai+m));        printf("%.3lf\n",ans);    }    return 0 ;}