problem:
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which
minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Array Dynamic Programming
题意:从矩阵的左上方到右下方寻找一条加权最小的路径
thinking:
(1)矩阵的路径问题(求路径总数。带障碍物的路径总数。加权最小、最大路径等),因为具有清晰且简易的状态转移公式,
统统能够用DP法解决,时间复杂度都为O(m*n)
(2) 用DP解决全局最优问题!
。该题是否具有局部最优解呢。状态转移公式:a[i][j] = min(a[i-1][j], a[i][j-1]) + grid[i][j];
规定了每一步的选择都是最优解。所以局部最优是成立的
code:
class Solution { public: int minPathSum(vector<vector<int> > &grid) { vector<vector<int> >::const_iterator con_it=grid.begin(); int m=grid.size(); int n=(*con_it).size(); vector<int> tmp(n,0); vector<vector<int> > a(m,tmp); if(m==0 || n==0) return 0; a[0][0]=grid[0][0]; for(int i=1;i<m;i++) a[i][0]=a[i-1][0]+grid[i][0]; for(int j=1;j<n;j++) a[0][j]=a[0][j-1]+grid[0][j]; for(int i = 1; i < m; i++) for(int j = 1; j < n; j++) a[i][j] = min(a[i-1][j], a[i][j-1]) + grid[i][j]; return a[m-1][n-1]; } };