[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：

#include <cstring>
#include <cmath>
class Solution {
public:
    /**
     * @param obstacleGrid: A list of lists of integers
     * @return: An integer
     */
    int solute(vector<vector<int> > &v, vector<vector<int> > &f, int m, int n)
    {
        if(f[m][n]!=-1)   return f[m][n];
        int M = v.size(), N = v[0].size();
        if(v[M-m][N-n]==1)  f[m][n] = 0;
        else if(m==1&&n==1) f[m][n] = 1;
        else if(m==1)    f[m][n] = min(1, solute(v, f, 1, n-1));
        else if(n==1)    f[m][n] = min(1, solute(v, f, m-1, 1));
        else    f[m][n] = solute(v, f, m-1, n)+solute(v, f, m, n-1);
        return f[m][n];
    }
    int uniquePathsWithObstacles(vector<vector<int> > &v) {
        // write your code here
        int m = v.size(), n = v[0].size();
        vector<vector<int> > f(m+1, vector<int>(n+1, -1));
        return solute(v, f, m, n);
    }
};