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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<vector>
 5 #include<cstring>
 6 #include<queue>
 7 #include<cmath>
 8 const double PI = acos(-1.0);//记住PI
 9 using namespace std;
10 int n, vis[10];
11 double maxm;
12 double x1, x2, yy1, y2, z1, z2;
13 double a[10], b[10], c[10], r[10];
14 double dist(double a, double b, double c){
15     return sqrt(a*a+b*b+c*c);
16 }
17 double cal_min(double a, double b, double c){
18     double tmp1 = min(fabs(a-x1), min(fabs(b-yy1), fabs(c-z1)));
19     double tmp2 = min(fabs(a-x2), min(fabs(b-y2), fabs(c-z2)));
20     return min(tmp1, tmp2);
21 }
22 void dfs(double sum, int k)
23 {
24     if(k == n){
25         maxm = max(sum, maxm);
26     }
27     else
28     for(int i = 0; i < n; i++){
29         if(!vis[i]){//主要问题处在下面的处理
30             r[i] = cal_min(a[i], b[i], c[i]);
31             for(int j = 0; j < n; j++){//这里不能加j!=i
32                 if(vis[j] == 1){
33                     if(dist(a[i]-a[j], b[i]-b[j], c[i]-c[j])>r[j])
34                         r[i] = min(dist(a[i]-a[j], b[i]-b[j], c[i]-c[j])-r[j], r[i]);
35                     else r[i] = 0;
36                 }
37             }
38             sum += r[i]*r[i]*r[i];
39             vis[i] = 1;
40             dfs(sum, k+1);
41             sum -= r[i]*r[i]*r[i];
42             vis[i] = 0;
43         }
44     }
45 }
46 int main()
47 {
48     int kase=0;
49     while(~scanf("%d", &n)&&n){
50
51         scanf("%lf%lf%lf", &x1, &yy1, &z1);
52         scanf("%lf%lf%lf", &x2, &y2, &z2);
53         for(int i = 0; i < n; i++){
54             scanf("%lf%lf%lf", &a[i], &b[i], &c[i]);
55         }
56         memset(vis, 0, sizeof(vis));
57         memset(r, 0, sizeof(r));
58         maxm = 0;
59         dfs(0, 0);
60         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
61     }
62     return 0;
63 }

原文地址：https://www.cnblogs.com/Surprisezang/p/10222770.html