蓝桥杯-PREV28-地宫取宝

先自己用dp解了一遍，然后看了官方讲解视频是用记忆化搜索做的。感觉那位老师的方法比较容易实现(效率上和我的差不多的)；记录一下三种方法。

• 动态规划
  

  地宫取宝

  1.195KB

  C++

  正确

  100

  15ms

  1.113MB

  #include "bits/stdc++.h"
  using namespace std;
  const int MOD = 1000000007;
  // mp是地宫地图，table[i][j]表示从坐标(1, 1)走到(i, j)的方案数 
  int mp[55][55], table[55][55];
  // dp[i][j][k]表示在坐标(i, j)手中有k件物品的方案数 
  int dp[55][55][15];
  int main() {
      int n, m, k;
      scanf("%d%d%d", &n, &m, &k);
      table[1][1] = 1;
      for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++)
      if (i != 1 || j != 1)
      table[i][j] = table[i - 1][j] + table[i][j - 1];
      for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++)
      scanf("%d", &mp[i][j]);
      for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
          dp[i][j][1] = table[i][j];
          for (int x = i; x <= n; x++)
          for (int y = j; y <= m; y++) {
              // dp[x][y][z]是从dp[i][j][z - 1]推出来的；
              // 要乘(i, j)到(x, y)的方案数; 把(i, j)看做(1, 1)就相当于乘table[x - i + 1][y - j + 1] 
              for (int z = 2; z <= k; z++)
              if (mp[x][y] > mp[i][j]) 
              dp[x][y][z] = (dp[x][y][z] + dp[i][j][z - 1] * 1LL * table[x - i + 1][y - j + 1]) % MOD ;
          }
      }
      int res = 0;
      for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++)  
      res = (res + dp[i][j][k] * 1LL * table[n - i + 1][m - j + 1]) % MOD;
      printf("%d
  ", res);
      return 0;
  } 

• 深度优先搜索

  地宫取宝

  734B

  C++

  运行超时

  42

  运行超时

  952.0KB

  #include "bits/stdc++.h"
  using namespace std;
  const int MOD = 1000000007;
  int mp[55][55];
  int n, m, k;
  int dfs(int x, int y, int mx, int cnt) {
      if (x > n || y > m || cnt > k) {
          return 0;
      }
      if (x == n && y == m) {
          if (cnt == k || (cnt == k - 1 && mx < mp[x][y])) {
              return 1;
          }
          return 0;
      }
      int res = 0;
      if (mp[x][y] > mx) {
          res = (res + dfs(x + 1, y, mp[x][y], cnt + 1)) % MOD;
          res = (res + dfs(x, y + 1, mp[x][y], cnt + 1)) % MOD;
      }
      res = (res + dfs(x + 1, y, mx, cnt)) % MOD;
      res = (res + dfs(x, y + 1, mx, cnt)) % MOD;
      return res;
  }
  int main() {
      scanf("%d%d%d", &n, &m, &k);
      for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++)
      scanf("%d", &mp[i][j]);
      printf("%d
  ", dfs(1, 1, -1, 0));
      return 0;
  }

   七组数据过了三组

• 记忆化搜索
  

  地宫取宝

  0.995KB

  C++

  正确

  100

  0ms

  3.535MB

  #include "bits/stdc++.h"
  using namespace std;
  const int MOD = 1000000007;
  int mp[55][55];
  int dp[55][55][15][15];
  int n, m, k;
  int dfs(int x, int y, int mx, int cnt) {
      if (x > n || y > m || cnt > k) {
          return 0;
      }
      // 因为mx可能为-1,所以这里mx + 1,后面的mx + 1也是一样 
      if (dp[x][y][mx + 1][cnt] != -1) {
          return dp[x][y][mx + 1][cnt];
      }
      if (x == n && y == m) {
          if (cnt == k || (cnt == k - 1 && mx < mp[x][y])) {
              dp[x][y][mx + 1][cnt] = 1;
              return 1;
          }
          dp[x][y][mx + 1][cnt] = 0;
          return 0;
      }
      int res = 0;
      if (mp[x][y] > mx) {
          res = (res + dfs(x + 1, y, mp[x][y], cnt + 1)) % MOD;
          res = (res + dfs(x, y + 1, mp[x][y], cnt + 1)) % MOD;
      }
      res = (res + dfs(x + 1, y, mx, cnt)) % MOD;
      res = (res + dfs(x, y + 1, mx, cnt)) % MOD;
      dp[x][y][mx + 1][cnt] = res;
      return res;
  }
  int main() {
      scanf("%d%d%d", &n, &m, &k);
      for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++)
      scanf("%d", &mp[i][j]);
      memset(dp, -1, sizeof(dp));
      printf("%d
  ", dfs(1, 1, -1, 0));
      return 0;
  }