UVALive 2474 Balloons in a Box（枚举）

https://vjudge.net/contest/277824#problem/A

尤其是模拟题，三思而后敲！！！

纠错了好久，主要还是没有处理好：单点若还未放气球，其他气球可以膨胀越过它（即可以无视这个点）；如果选到一个点，它已经在某一气球半径内了，则置r=0.

还参考了一些其他测试数据

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<vector>
 #include<cstring>
 #include<queue>
 #include<cmath>
 const double PI = acos(-1.0);//记住PI
 using namespace std;
 int n, vis[];
 double maxm;
 double x1, x2, yy1, y2, z1, z2;
 double a[], b[], c[], r[];
 double dist(double a, double b, double c){
     return sqrt(a*a+b*b+c*c);
 }
 double cal_min(double a, double b, double c){
     double tmp1 = min(fabs(a-x1), min(fabs(b-yy1), fabs(c-z1)));
     double tmp2 = min(fabs(a-x2), min(fabs(b-y2), fabs(c-z2)));
     return min(tmp1, tmp2);
 }
 void dfs(double sum, int k)
 {
     if(k == n){
         maxm = max(sum, maxm);
     }
     else
     for(int i = ; i < n; i++){
         if(!vis[i]){//主要问题处在下面的处理
             r[i] = cal_min(a[i], b[i], c[i]);
             for(int j = ; j < n; j++){//这里不能加j!=i
                 if(vis[j] == ){
                     if(dist(a[i]-a[j], b[i]-b[j], c[i]-c[j])>r[j])
                         r[i] = min(dist(a[i]-a[j], b[i]-b[j], c[i]-c[j])-r[j], r[i]);
                     else r[i] = ;
                 }
             }
             sum += r[i]*r[i]*r[i];
             vis[i] = ;
             dfs(sum, k+);
             sum -= r[i]*r[i]*r[i];
             vis[i] = ;
         }
     }
 }
 int main()
 {
     int kase=;
     while(~scanf("%d", &n)&&n){

         scanf("%lf%lf%lf", &x1, &yy1, &z1);
         scanf("%lf%lf%lf", &x2, &y2, &z2);
         for(int i = ; i < n; i++){
             scanf("%lf%lf%lf", &a[i], &b[i], &c[i]);
         }
         memset(vis, , sizeof(vis));
         memset(r, , sizeof(r));
         maxm = ;
         dfs(, );
         printf("Box %d: %.0lf\n\n", ++kase, (fabs(x2-x1)*fabs(y2-yy1)*fabs(z2-z1)-maxm*PI*4.0/3.0));//a*b*c-(4/3)*PI*r*r*r
     }
     return ;
 }