1. 题意
矩阵中一个位置只能从左上一、左、左下一格子转移而来，且当前值一定大于转移之前的值；
求从第一列开始的最大转移步数。
矩阵中移动的最大次数
2. 题解
- 思路
由于状态只能从左向右转移，所以同一个位置被搜索到后，第一列其他位置再搜索到它的距离一定相等。题目求得是第一列转移到其他位置的最大次数，我们需要把第一列置0,其他列置
MINVAL。使得只有从第一列转移的才能使得值为正数。
class Solution {
public:struct Dir {constexpr static int dir[][2] = {{1,-1},{0,-1},{-1,-1}};};int maxMoves(vector<vector<int>>& grid) {int w = grid[0].size();int h = grid.size();int MIN_VAL = -2000;int ans = 0;vector<vector<int>> dp(h, vector<int>(w, MIN_VAL));for (int i = 0;i < h; ++i)dp[i][0] = 0;for (int i = 1; i < w; ++i) {for (int j = 0;j < h; ++j) {for (auto &d:Dir::dir){int px = j + d[0];int py = i + d[1];if (px < 0 || py < 0 || px > h - 1 || py > w - 1)continue; if (grid[px][py] >= grid[j][i])continue;dp[j][i] = max(dp[j][i], dp[px][py] + 1);}ans = max(dp[j][i],ans);}}return ans;}
};
- 记忆化搜索
class Solution {
public:struct Dir {constexpr static int dir[][2] = {{-1,1},{0,1},{1,1}};};int dfs(int i, int j,vector<vector<int>> &mem, const vector<vector<int>> &grid){int m = mem.size();int n = mem[0].size();if (mem[i][j] != -1)return mem[i][j];int ans = 0;for (auto c:Dir::dir) {int nx = i + c[0];int ny = j + c[1];if ( nx < 0 || nx > m - 1 || ny < 0 || ny > n - 1)continue;if (grid[i][j] >=grid[nx][ny])continue;int t = dfs(nx,ny, mem, grid);ans = max(ans, 1 + t);}return mem[i][j] = ans;}int maxMoves(vector<vector<int>>& grid) {int ans = 0;int h = grid.size();int w = grid[0].size();vector<vector<int>> mem(h, vector<int>(w, -1));for (int i = 0;i < h; ++i) {ans = max(dfs(i, 0, mem, grid), ans);}return ans;}
};
)
