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https://leetcode.com/problems/longest-increasing-path-in-a-matrix/
Given an integer matrix, find the length of the longest increasing path.
From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).
Example 1:
nums = [ [9,9,4], [6,6,8], [2,1,1] ]
Return 4
The longest increasing path is [1, 2, 6, 9].
Example 2:
nums = [ [3,4,5], [3,2,6], [2,2,1] ]
Return 4
The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.
class Solution { public: bool checkRange(vector<vector<int> >& matrix, int x, int y) { int n = matrix.size(), m = matrix[0].size(); if(x < 0 || y < 0 || x >= n || y >= m) return false; return true; } int dfs(vector<vector<int> >& matrix, vector<vector<int> >& dp, vector<vector<bool> >& vis, vector<vector<int> >& dir, int x, int y) { if(dp[x][y]) return dp[x][y]; int MAX = -1; for(int i=0; i<dir.size(); ++i) { int nx = x + dir[i][0], ny = y + dir[i][1]; if(checkRange(matrix, nx, ny) && !vis[nx][ny] && matrix[nx][ny] > matrix[x][y]) { vis[nx][ny] = true; MAX = max(MAX, dfs(matrix, dp, vis, dir, nx, ny) + 1); vis[nx][ny] = false; } } if(MAX == -1) return 1; dp[x][y] = MAX; return dp[x][y]; } int longestIncreasingPath(vector<vector<int>>& matrix) { int n = matrix.size(); if(n == 0) return 0; int m = matrix[0].size(); vector<vector<int> > dp(n, vector<int>(m, 0)); vector<vector<bool> > vis(n, vector<bool>(m, false)); vector<vector<int> > dir(4); dir[0].push_back(-1); dir[0].push_back(0); dir[1].push_back(0); dir[1].push_back(1); dir[2].push_back(1); dir[2].push_back(0); dir[3].push_back(0); dir[3].push_back(-1); int res = -1; for(int i=0; i<n; ++i) { for(int j=0; j<m; ++j) { vis[i][j] = true; dp[i][j] = dfs(matrix, dp, vis, dir, i, j); res = max(res, dp[i][j]); vis[i][j] = false; } } return res; } };
leetcode@ [329] Longest Increasing Path in a Matrix (DFS + 记忆化搜索)
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原文地址：http://www.cnblogs.com/fu11211129/p/5146874.html
