[LintCode]unique paths with obstacles

http://www.lintcode.com/zh-cn/problem/unique-paths-ii/

在上一题的基础上加入了障碍物。

同样可采用递归实现，递推公式不变，但是需要加入对障碍物的判断。

下面是实现的代码：

 1 #include <cstring>
 2 #include <cmath>
 3 class Solution {
 4 public:
 5     /**
 6      * @param obstacleGrid: A list of lists of integers
 7      * @return: An integer
 8      */
 9     int solute(vector<vector<int> > &v, vector<vector<int> > &f, int m, int n)
10     {
11         if(f[m][n]!=-1)   return f[m][n];
12         int M = v.size(), N = v[0].size();
13         if(v[M-m][N-n]==1)  f[m][n] = 0;
14         else if(m==1&&n==1) f[m][n] = 1;
15         else if(m==1)    f[m][n] = min(1, solute(v, f, 1, n-1));
16         else if(n==1)    f[m][n] = min(1, solute(v, f, m-1, 1));
17         else    f[m][n] = solute(v, f, m-1, n)+solute(v, f, m, n-1);
18         return f[m][n];
19     }
20     int uniquePathsWithObstacles(vector<vector<int> > &v) {
21         // write your code here
22         int m = v.size(), n = v[0].size();
23         vector<vector<int> > f(m+1, vector<int>(n+1, -1));
24         return solute(v, f, m, n);
25     }
26 };