10012 - How Big Is It?

描述：好麻烦的题目，准确的说就是给定m个圆的半径，现在要求找到一个矩形使得每一个球都以地面相切，要求输出最小的矩阵的长度#include <cstdio>#include <cmath>#include <cstring>int n,m,j,flag[12];double count,num[10],p[10],value[10];void dfs(int cur,double x,double y){    if(cur>=m)    {        if(!flag[10]) count=y-x;        else if(y-x<count) count=y-x;        flag[10]=1;        return ;    }    for(int i=0; i<m; i++)        if(!flag[i])        {            flag[i]=1;            p[cur]=num[i];            value[cur]=0;            for (j=0; j<cur; j++)            {                double c=sqrt((p[cur]+p[j])*(p[cur]+p[j])-(p[cur]-p[j])*(p[cur]-p[j]))+value[j];                if (c>value[cur]) value[cur]=c;            }            double a,b;            if(value[cur]-p[cur]<x) a=value[cur]-p[cur];            else a=x;            if(value[cur]+p[cur]>y) b=value[cur]+p[cur];            else b=y;            dfs(cur+1,a,b);            flag[i]=0;        }}int main(){#ifndef ONLINE_JUDGE    freopen("a.txt","r",stdin);#endif    scanf("%d",&n);    while(n--)    {        scanf("%d",&m);        count=0;        for(int i=0; i<m; i++) scanf("%lf",&num[i]);        memset(flag,0,sizeof(flag));        memset(value,0,sizeof(value));        if(m>1) for(int i=0; i<m; i++)            {                flag[i]=1;                value[0]=p[0]=num[i];                double a=value[0]-p[0];                double b=value[0]+p[0];                dfs(1,a,b);                flag[i]=0;            }        else count=2*num[0];        printf("%.3lf\n",count);    }    return 0;}