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]
]

The total number of unique paths is

2

.

Note: m and n will be at most 100.

思路：这也是一道DP的题，在unique paths上设置了路径障碍。可用数组ob[m]
=-1表示在点(m,n)有障碍不能通过，则到达该点的路径数f[m]
=0。边界条件是当m=0||n=0时，f[m]
=0，当m=1&&n=1&&ob[m]
!=-1时，才能到达点(m,n)有一条路径，状态转移公式同I。

class Solution {
private:
int ob[101][101];
int f[101][101];
public:
int unique(int m, int n)
{
if (ob[m]
== -1)
{
return 0;
}
if (m==0 || n==0)
{
return 0;
}
if (m == 1 && n == 1 && ob[m]
!= -1)
{
f[m]
= 1;
return 1;
}
if (ob[m]
!= -1 && f[m]
!= 0)
{
return f[m]
;
}
f[m]
= unique(m-1,n) + unique(m, n-1);
return f[m]
;
}
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
memset(ob, 0, sizeof(ob));
memset(f, 0, sizeof(f));
int m = obstacleGrid.size(),i,j;
if (m == 0)
{
return 0;
}
int n = obstacleGrid[0].size();
for(i=0; i<m; ++i)
for(j=0; j<n; ++j)
if(obstacleGrid[i][j] == 1)
{
ob[i+1][j+1] = -1;
}
return unique(m, n);
}
};