之所以将这三道题放在一起，是因为这三道题非常类似。
1. Minimum Path Sum
题目要求：
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.
该题解答参考自一博文。
设dp[i][j]表示从左上角到grid[i][j]的最小路径和。那么dp[i][j] = grid[i][j] + min( dp[i-1][j], dp[i][j-1] );
下面的代码中，为了处理计算第一行和第一列的边界条件，我们令dp[i][j]表示从左上角到grid[i-1][j-1]的最小路径和，最后dp[rows][cols]是我们所求的结果
1 int minPathSum(vector<vector<int>>& grid) { 2 int rows = grid.size(); 3 if(rows == 0) 4 return 0; 5 int cols = grid[0].size(); 6 7 vector<vector<int> > dp(rows + 1, vector<int>(cols + 1, INT_MAX)); 8 dp[0][1] = 0; 9 for(int i = 1; i < rows + 1; i++) 10 for(int j = 1; j < cols + 1; j++) 11 dp[i][j] = grid[i - 1][j - 1] + min(dp[i][j - 1], dp[i - 1][j]); 12 13 return dp[rows][cols]; 14 }