这里会给出关于动规的一些常见题型和变种，给自己做一个参考：
Leetcode #62. Unique Paths 路径搜寻 解题报告
1、题目
A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).
How many possible unique paths are there?

public class Solution { 
 public int uniquePaths(int m, int n) {
   int[][] dp=new int[m][n];
   int i,j;
   for(i=0;i<n;i++)
     dp[0][i]=1;
   for(i=0;i<m;i++) 
     dp[i][0]=1;
  for(i=1;i<m;i++){ 
  for(j=1;j<n;j++){ 
     dp[i][j]=dp[i-1][j]+dp[i][j-1]; } } 
return dp[m-1][n-1];
 }}

思路2.采用排列组合的思想
对于m*n的格子，一样的，就是从m+n步中选出m步向下或n步向右，因此为C(m+n,m)=C(m+n,n)种。

变形1.63. Unique Paths II
Follow up for "Unique Paths":
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1
and 0
respectively in the grid.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

这道题相比较上一道题，有的地方是不能走的，使用对应矩阵当中1来标示。
与上题类似,如果不能走，A[i][j] = 0，因此，只需要加一个判读条件就好。
代码：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m=obstacleGrid.size();
        int n=obstacleGrid[0].size();
        vector<vector<int>> dp(m,vector<int>(n,0));
        int i=0,j;

        while(i<n && obstacleGrid[0][i]==0){
            dp[0][i++]=1;
        }
        i=0;
        while(i<m && obstacleGrid[i][0]==0){
            dp[i++][0]=1;
        }
        for(i=1;i<m;i++){
            for(j=1;j<n;j++){
                if(obstacleGrid[i][j]==0)
                    dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
};

变形3：Leetcode 64. Minimum Path
题目：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.
（感觉只要是只能向下和向右走的都可以用这种思想啊）
代码：

public class Solution {
    public int minPathSum(int[][] grid) {
        if(grid.length==0 || grid[0].length==0)
            return 0;
        int m=grid.length;
        int n=grid[0].length;
        int dp[][]=new int[m][n];
        int top,left;
        int i=0,j;
        dp[0][0]=grid[0][0];
        for( j=1;j<n;j++){
            dp[i][j]=dp[i][j-1]+grid[i][j];
        }
         for( j=1;j<m;j++){
            dp[j][i]=dp[j-1][i]+grid[j][i];
        }
        for( i=1;i<m;i++){
            for( j=1;j<n;j++){
                top=Integer.MAX_VALUE;
                left=Integer.MAX_VALUE;
                top=dp[i-1][j];
                left=dp[i][j-1];
                dp[i][j]=Math.min(top,left)+grid[i][j];
            }
        }
        return dp[m-1][n-1];
    }
}