题目描述：设想有个机器人坐在X*Y网格的左上角，只能向右向下移动。机器人从(0,0)开始出发，到(X,Y)共有多少种方法。

思路：到i，j只和，(i-1,j)和(i,j-1)有关

递归的时候加备忘

 1 #include <iostream>
 2 #include <queue>
 3 #include <climits>
 4 #include <algorithm>
 5 #include <memory.h>
 6 #include <stdio.h>
 7 #include <ostream>
 8 #include <vector>
 9 #include <list>
10 #include <cmath>
11 #include <string>
12 #include <stdexcept>
13 #include <stack>
14 using namespace std;
15 
16 int gridPaths[1000][1000];
17 
18 int fun(int x,int y)
19 {
20     if(x == 0 || y == 0)
21     {
22         gridPaths[x][y] = 1;
23         return gridPaths[x][y];
24     }
25     if(gridPaths[x][y] > 0)
26         return gridPaths[x][y];
27     else
28     {
29         gridPaths[x][y] = fun(x-1,y)+fun(x,y-1);
30         return gridPaths[x][y];
31     }
32 }
33 
34 int main()
35 {
36     memset(gridPaths,-1,sizeof(gridPaths));
37     int x = 10;
38     int y = 10;
39     cout<<fun(x,y);
40     return 0;
41 }

思考：如果加入禁区点，怎么找一条路径？