首先分析题目：
"对(x, y, z)的立方体消毒，cost = min(x, y, z)"
所以由贪心的想法，我们可以对(1, y, z)的立方体消毒，cost = 1，这一定是最优的cost为1的解法
又a * b * c ≤ 5000, 不妨设a = min(a, b, c) ≤ 17，故我们可以枚举a方向哪些(1, b, c)的立方体被消毒了。
然后把剩下的部分压成一个(b, c)的矩形，于是就变成经典问题最小点覆盖了，转化成二分图最大匹配来做。
1 /************************************************************** 2 Problem: 3140 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:1324 ms 7 Memory:1312 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 #include <cstring> 13 14 using namespace std; 15 const int N = 5005; 16 17 int a, b, c, ans; 18 int first[N], tot, cnt_p, now_time; 19 int l[N], vis[N]; 20 bool u[20]; 21 22 struct point { 23 int x, y, z; 24 point() {} 25 point(int _x, int _y, int _z) : x(_x), y(_y), z(_z) { 26 if (b <= a && b <= c) swap(x, y); else 27 if (c <= a && c <= b) swap(x, z); 28 } 29 } p[N]; 30 31 struct edge { 32 int next, to; 33 edge() {} 34 edge(int _n, int _t) : next(_n), to(_t) {} 35 } e[N * 10]; 36 37 int read() { 38 int x = 0; 39 char ch = getchar(); 40 while (ch < '0' || '9' < ch) 41 ch = getchar(); 42 while ('0' <= ch && ch <= '9') 43 (x *= 10) += ch - '0', ch = getchar(); 44 return x; 45 } 46 47 inline void add_edge(int x, int y) { 48 e[++tot] = edge(first[x], y), first[x] = tot; 49 } 50 51 #define y e[x].to 52 bool find(int p) { 53 int x; 54 for (x = first[p]; x; x = e[x].next) 55 if (vis[y] != now_time) { 56 vis[y] = now_time; 57 if (!l[y] || find(l[y])) { 58 l[y] = p; 59 return 1; 60 } 61 } 62 return 0; 63 } 64 #undef y 65 66 inline void work(int res) { 67 int i; 68 tot = 0; 69 for (i = 1; i <= b; ++i) first[i] = 0; 70 for (i = 1; i <= c; ++i) l[i + b] = 0; 71 for (i = 1; i <= cnt_p; ++i) 72 if (!u[p[i].x]) 73 add_edge(p[i].y, p[i].z + b); 74 for (i = 1; i <= b; ++i) { 75 ++now_time; 76 if ((res += find(i)) > ans) return; 77 } 78 ans = res; 79 } 80 81 void dfs(int step, int now_ans) { 82 if (now_ans > ans) return; 83 if (step > a) { 84 work(now_ans); 85 return; 86 } 87 u[step] = 1, dfs(step + 1, now_ans + 1); 88 u[step] = 0, dfs(step + 1, now_ans); 89 } 90 91 int main() { 92 int i, j, k, T; 93 T = read(); 94 while (T--) { 95 a = read(), b = read(), c = read(), cnt_p = 0; 96 for (i = 1; i <= a; ++i) 97 for (j = 1; j <= b; ++j) 98 for (k = 1; k <= c; ++k) 99 if (read()) p[++cnt_p] = point(i, j, k); 100 if (b <= a && b <= c) swap(a, b); else 101 if (c <= a && c <= b) swap(a, c); 102 ans = a; 103 dfs(1, 0); 104 printf("%d ", ans); 105 } 106 return 0; 107 }