BZOJ3140 [Hnoi2013]消毒

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\n", ans);
105   }
106   return 0;
107 }
View Code

 

posted on 2015-01-28 17:56  Xs酱~  阅读(221)  评论(0)  编辑  收藏  举报