leetcode || 63、Unique Paths II

problem：

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]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

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Array Dynamic
Programming

题意：紧挨着上一题，只不过对矩阵作了一点改变：矩阵的0代表可以通行，1代表不可通行，

求从起点（0,0）到终点（m-1,n-1）的路径总数

thinking：

（1）blog.csdn.net/hustyangju/article/details/44829339 讨论了只能使用DP法

（2）加了条件限制，则DP算法也要修改：

1、边界条件要改变，一旦出现1，则随后的边界条件全部为0

2、矩阵中间出现1，则该位置的路径数置为0

code：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
    vector<vector<int> >::const_iterator con_it=obstacleGrid.begin();
    int m=obstacleGrid.size();
    int n=(*con_it).size();
    vector<int> tmp(n,0);
    vector<vector<int> > a(m,tmp);
     bool flag=true;
    for(int i=0;i<m;i++)  //边界条件
    {
        if(obstacleGrid[i][0]==0 && flag)
            a[i][0]=1;
        else
        {
            a[i][0]=0;
            flag=false;
        }

    }
    flag=true;
    for(int j=0;j<n;j++) //边界条件
    {
        if(obstacleGrid[0][j]==0 && flag)
            a[0][j]=1;
        else
        {
            a[0][j]=0;
            flag=false;
        }
    }
    for(int i = 1; i < m; i++)
        for(int j = 1; j < n; j++)
        {
            if(obstacleGrid[i][j]==1)  //出现障碍物，置0
                a[i][j]=0;
            else
                a[i][j] = a[i-1][j] + a[i][j-1];
        }
    return a[m-1][n-1];
    }
};