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 = ;
  // 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[][], table[][];
  // dp[i][j][k]表示在坐标(i, j)手中有k件物品的方案数
  int dp[][][];
  int main() {
      int n, m, k;
      scanf("%d%d%d", &n, &m, &k);
      table[][] = ;
      for (int i = ; i <= n; i++)
      for (int j = ; j <= m; j++)
      if (i !=  || j != )
      table[i][j] = table[i - ][j] + table[i][j - ];
      for (int i = ; i <= n; i++)
      for (int j = ; j <= m; j++)
      scanf("%d", &mp[i][j]);
      for (int i = ; i <= n; i++)
      for (int j = ; j <= m; j++) {
          dp[i][j][] = 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 = ; z <= k; z++)
              if (mp[x][y] > mp[i][j])
              dp[x][y][z] = (dp[x][y][z] + dp[i][j][z - ] * 1LL * table[x - i + ][y - j + ]) % MOD ;
          }
      }
      int res = ;
      for (int i = ; i <= n; i++)
      for (int j = ; j <= m; j++)
      res = (res + dp[i][j][k] * 1LL * table[n - i + ][m - j + ]) % MOD;
      printf("%d\n", res);
      return ;
  } 

• 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 = ;
  int mp[][];
  int n, m, k;
  int dfs(int x, int y, int mx, int cnt) {
      if (x > n || y > m || cnt > k) {
          return ;
      }
      if (x == n && y == m) {
          if (cnt == k || (cnt == k -  && mx < mp[x][y])) {
              return ;
          }
          return ;
      }
      int res = ;
      if (mp[x][y] > mx) {
          res = (res + dfs(x + , y, mp[x][y], cnt + )) % MOD;
          res = (res + dfs(x, y + , mp[x][y], cnt + )) % MOD;
      }
      res = (res + dfs(x + , y, mx, cnt)) % MOD;
      res = (res + dfs(x, y + , mx, cnt)) % MOD;
      return res;
  }
  int main() {
      scanf("%d%d%d", &n, &m, &k);
      for (int i = ; i <= n; i++)
      for (int j = ; j <= m; j++)
      scanf("%d", &mp[i][j]);
      printf("%d\n", dfs(, , -, ));
      return ;
  }

  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 = ;
  int mp[][];
  int dp[][][][];
  int n, m, k;
  int dfs(int x, int y, int mx, int cnt) {
      if (x > n || y > m || cnt > k) {
          return ;
      }
      // 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 + ][cnt] != -) {
          return dp[x][y][mx + ][cnt];
      }
      if (x == n && y == m) {
          if (cnt == k || (cnt == k -  && mx < mp[x][y])) {
              dp[x][y][mx + ][cnt] = ;
              return ;
          }
          dp[x][y][mx + ][cnt] = ;
          return ;
      }
      int res = ;
      if (mp[x][y] > mx) {
          res = (res + dfs(x + , y, mp[x][y], cnt + )) % MOD;
          res = (res + dfs(x, y + , mp[x][y], cnt + )) % MOD;
      }
      res = (res + dfs(x + , y, mx, cnt)) % MOD;
      res = (res + dfs(x, y + , mx, cnt)) % MOD;
      dp[x][y][mx + ][cnt] = res;
      return res;
  }
  int main() {
      scanf("%d%d%d", &n, &m, &k);
      for (int i = ; i <= n; i++)
      for (int j = ; j <= m; j++)
      scanf("%d", &mp[i][j]);
      memset(dp, -, sizeof(dp));
      printf("%d\n", dfs(, , -, ));
      return ;
  }