方法：暴力 枚举
数据量较小，可以枚举所有n!个order，然后依次计算该order所对应的体积，更新答案。因为没有剪枝，所以用next_permutation 列出所有可能性即可。
code:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <string> 6 #include <vector> 7 #include <stack> 8 #include <bitset> 9 #include <cstdlib> 10 #include <cmath> 11 #include <set> 12 #include <list> 13 #include <deque> 14 #include <map> 15 #include <queue> 16 #include <fstream> 17 #include <cassert> 18 #include <unordered_map> 19 #include <unordered_set> 20 #include <cmath> 21 #include <sstream> 22 #include <time.h> 23 #include <complex> 24 #include <iomanip> 25 #define Max(a,b) ((a)>(b)?(a):(b)) 26 #define Min(a,b) ((a)<(b)?(a):(b)) 27 #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a)) 28 #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a)) 29 #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a)) 30 #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a)) 31 #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a)) 32 #define FOREACH(a,b) for (auto &(a) : (b)) 33 #define rep(i,n) FOR(i,0,n) 34 #define repn(i,n) FORN(i,1,n) 35 #define drep(i,n) DFOR(i,n-1,0) 36 #define drepn(i,n) DFOR(i,n,1) 37 #define MAX(a,b) a = Max(a,b) 38 #define MIN(a,b) a = Min(a,b) 39 #define SQR(x) ((LL)(x) * (x)) 40 #define Reset(a,b) memset(a,b,sizeof(a)) 41 #define fi first 42 #define se second 43 #define mp make_pair 44 #define pb push_back 45 #define all(v) v.begin(),v.end() 46 #define ALLA(arr,sz) arr,arr+sz 47 #define SIZE(v) (int)v.size() 48 #define SORT(v) sort(all(v)) 49 #define REVERSE(v) reverse(ALL(v)) 50 #define SORTA(arr,sz) sort(ALLA(arr,sz)) 51 #define REVERSEA(arr,sz) reverse(ALLA(arr,sz)) 52 #define PERMUTE next_permutation 53 #define TC(t) while(t--) 54 #define forever for(;;) 55 #define PINF 1000000000000 56 #define newline ' ' 57 58 #define test if(1)if(0)cerr 59 using namespace std; 60 using namespace std; 61 typedef vector<int> vi; 62 typedef vector<vi> vvi; 63 typedef pair<int,int> ii; 64 typedef pair<double,double> dd; 65 typedef pair<char,char> cc; 66 typedef vector<ii> vii; 67 typedef long long ll; 68 typedef unsigned long long ull; 69 typedef pair<ll, ll> l4; 70 const double pi = acos(-1.0); 71 72 struct Point 73 { 74 double x[3]; 75 }; 76 double distance(const Point &l, const Point &r) 77 { 78 double ret = 0; 79 rep(i, 3) 80 { 81 double d = l.x[i]-r.x[i]; 82 ret += d*d; 83 } 84 return sqrt(ret); 85 } 86 Point v[2]; 87 Point center[6]; 88 double r[6]; 89 int perm[6] = {0,1,2,3,4,5}; 90 int n; 91 double ans; 92 void dfs(int pos, double vol) 93 { 94 int cur = perm[pos]; 95 if (pos == n) 96 { 97 ans = max(ans, vol); 98 return; 99 } 100 double tmp = abs(v[0].x[0]-center[cur].x[0]); 101 rep(i, 2) rep(j, 3) 102 tmp = min(tmp, abs(v[i].x[j]-center[cur].x[j])); 103 rep(i, pos) 104 { 105 if (!r[perm[i]]) continue; 106 tmp = min(tmp, distance(center[perm[i]], center[cur])-r[perm[i]]); 107 } 108 if (tmp <= 0) 109 { 110 r[cur] = 0; 111 } 112 else 113 r[cur] = tmp; 114 dfs(pos+1, vol + pi * 4 / 3 * r[cur]*r[cur]*r[cur]); 115 } 116 int main() 117 { 118 ios::sync_with_stdio(false); 119 cin.tie(0); 120 int kase = 0; 121 while (cin >> n && n) 122 { 123 124 rep(i, 2) 125 rep(j, 3) cin >> v[i].x[j]; 126 rep(i, n) 127 rep(j, 3) cin >> center[i].x[j]; 128 ans = 0; 129 do 130 { 131 Reset(r, 0); 132 dfs(0, 0); 133 } while (next_permutation(perm, perm+n)); 134 double volume = 1; 135 rep(i, 3) volume *= v[1].x[i]-v[0].x[i]; 136 volume = abs(volume); 137 ans = volume-ans; 138 cout << "Box " << ++kase << ": " << (int) (ans + 0.5) << newline << newline;; 139 } 140 141 } 142 143 /* 144 2 145 0 0 0 146 10 10 10 147 3 3 3 148 7 7 7 149 0 150 */