No Abstract.

Description

有$n \times m$顆巧克力, 排成矩陣, 每顆有一個顏色和價值, 請你選出盡量少的巧克力, 使得至少包含$k$種不同顏色, 在此基礎上, 使價值中位數最小.

Solution

這什麼神仙解法啊……

首先, 如果拋開中位數的限制, 這就是一個斯坦納樹. 但是顏色很不好搞, 因爲顏色有點多, 考慮到用上的不多, 我們對每個顏色隨機地做$[1,k]$的映射, 求包含五種狀態的斯坦納數. 這樣是有$\frac{5!}{5^5}$的正確率的, 所以做那麼一百多輪斯坦納樹就好啦.

然後考慮二分中位數. 這裏有個奇技淫巧的說… 把小於等於當前答案的$val$都設爲$999$, 反之是$1001$. 因爲點數最多$233$個, 所以我們放大的範圍的一半還是比點數多, 這樣上取整就不會擔心誤差啦. 實際的點數就是$\frac{ans + 500}{1000}$. 然後如果二分的答案小於點數 $\times 1000$的話, 就說明答案還可以更小. 繼續二分過去就好了.

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#include <bits/stdc++.h>

using namespace std;
 
const int maxn = 250;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

int n, m, k, col[maxn], ori[maxn][maxn], orv[maxn][maxn];
int f[maxn][70], val[maxn];
int mp[maxn];

inline int rd() {
  register int x = 0, f = 0, c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + (c ^ 48), c = getchar();
  return f ? -x : x;
}
struct edge {
  int to, nxt;
} e[maxn << 3];
int ptr, lnk[maxn], vis[maxn];
int id[maxn][maxn], cnt, tmp;
queue<int> q;

inline void add(int bgn, int end) {
  e[++ptr] = (edge){end, lnk[bgn]};
  lnk[bgn] = ptr;
}
inline void spfa(int s) {
  while (!q.empty()) {
    int u = q.front(); q.pop(); vis[u] = 0;
    for (int p = lnk[u]; p; p = e[p].nxt) {
      int y = e[p].to;
      if (!(~col[y])) continue;
      if (f[y][s] > f[u][s] + val[y]) {
        f[y][s] = f[u][s] + val[y];
        if (!vis[y]) 
          vis[y] = 1, q.push(y);
      }
    }
  }
}
inline void solve() {
  //queue<int> q;
  memset(f, 0x3f, sizeof f);
  for (int i = 1; i <= n * m; ++i)
    if (~col[i])
      f[i][1 << col[i]] = val[i];
  int lim = (1 << k) - 1;
  for (int s = 1; s <= lim; ++s) {
    for (int i = 1; i <= n * m; ++i)
      for (int sn = s - 1; sn; sn = (sn - 1) & s)
        f[i][s] = min(f[i][s], f[i][s ^ sn] + f[i][sn] - val[i]);
    for (int i = 1; i <= n * m; ++i)
      if (f[i][s] != 0x3f3f3f3f) 
        vis[i] = 1, q.push(i);
    spfa(s);
  }
  for (int i = 1; i <= n * m; ++i) 
    tmp = min(tmp, f[i][lim]);
}

int main() {
  srand(time(NULL));
  int T = rd();
  while (T--) {
    ptr = 0; cnt = 0;
    memset(lnk, 0, sizeof lnk);
    n = rd(); m = rd(); k = rd();
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j)
        id[i][j] = ++cnt;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j)
        ori[i][j] = rd();
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j)
        orv[i][j] = rd();
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j) 
        for (int k = 0; k < 4; ++k) {
          int ix = i + dx[k], iy = j + dy[k];
          if (ix >= 1 && ix <= n && iy >= 1 && iy <= m)
            add(id[i][j], id[ix][iy]);
        }
    int l = 0, r = 1e+6 + 5, pnt;
    bool nosolu = false;
    while (l <= r) {
      //random
      int mid = (l + r) >> 1, ans = 0x3f3f3f3f;
      for (int CNT_ = 1; CNT_ < 233; ++CNT_) {
        tmp = 0x3f3f3f3f;
        for (int i = 1; i <= n * m; ++i)
          mp[i] = rand() % k;
        for (int i = 1; i <= n; ++i)
          for (int j = 1; j <= m; ++j)
            col[id[i][j]] = (ori[i][j] == -1 ? -1 : mp[ori[i][j]]),
            val[id[i][j]] = (orv[i][j] <= mid ? 999 : 1001);
        solve();
        ans = min(ans, tmp);
      }
      if (ans == 0x3f3f3f3f) {
        nosolu = true;
        break;
      }
      int trans = (ans + 500) / 1000;
      if (ans <= trans * 1000) 
        r = mid - 1, pnt = trans;
      else 
        l = mid + 1;
    }
    if (nosolu) puts("-1 -1");
    else printf("%d %d\n", pnt, r + 1);
  }
  return 0;
}