题面

题解

当只有二维时，就是一个二分图匹配的板子题

三维的时候就很好做了，暴力枚举一维的情况，因为\(\min(x,y,z) = \sqrt{5000} < 18\)，于是时间复杂度有保证

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))inline int read()
{int data = 0, w = 1; char ch = getchar();while(ch != '-' && (!isdigit(ch))) ch = getchar();if(ch == '-') w = -1, ch = getchar();while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();return data * w;
}const int maxn(5010);
struct edge { int next, to; } e[maxn];
int head[maxn], e_num, a, b, c, Min, pos[4][maxn], vis[maxn], match[maxn], clean[maxn], ans, cnt, T;inline void add_edge(int from, int to) { e[++e_num] = (edge) {head[from], to}; head[from] = e_num; }
bool dfs(int x)
{for(RG int i = head[x]; i; i = e[i].next){int to = e[i].to; if(vis[to]) continue; vis[to] = true;if(!match[to] || dfs(match[to])) return match[to] = x, true;}return false;
}inline void Doit(int x)
{using std::fill; e_num = 0;fill(head + 1, head + b + 1, 0);fill(match + 1, match + c + 1, 0);fill(clean + 1, clean + a + 1, 1);int res = 0;for(RG int i = 0; i < a; i++)if(x & (1 << i)) clean[i + 1] = 0, ++res;for(RG int i = 1; i <= cnt; i++)if(clean[pos[1][i]]) add_edge(pos[2][i], pos[3][i]);for(RG int i = 1; i <= b; i++){fill(vis + 1, vis + c + 1, 0);if(dfs(i)) ++res;}ans = std::min(ans, res);
}int main()
{T = read();while(T--){cnt = 0; ans = 0x3f3f3f3f;a = read(); b = read(); c = read(); Min = std::min(std::min(a, b), c);for(RG int i = 1, x; i <= a; i++)for(RG int j = 1; j <= b; j++)for(RG int k = 1; k <= c; k++)if((x = read())) ++cnt, pos[1][cnt] = i, pos[2][cnt] = j, pos[3][cnt] = k;using std::swap;if(Min == b) swap(a, b), swap(pos[1], pos[2]);if(Min == c) swap(a, c), swap(pos[1], pos[3]);for(RG int i = 0; i < (1 << a); i++) Doit(i);printf("%d\n", ans);}return 0;
}