题目：
Description
A robot has to patrol around a rectangular area which is in a form of mxn grid (m rows and n columns). The rows are labeled from 1 to m. The columns are labeled from 1 to n. A cell (i, j) denotes the cell in row i and column j in the grid. At each step, the robot can only move from one cell to an adjacent cell, i.e. from (x, y) to (x + 1, y), (x, y + 1), (x - 1, y) or (x, y - 1). Some of the cells in the grid contain obstacles. In order to move to a cell containing obstacle, the robot has to switch to turbo mode. Therefore, the robot cannot move continuously to more than k cells containing obstacles.
Your task is to write a program to find the shortest path (with the minimum number of cells) from cell (1, 1) to cell (m, n). It is assumed that both these cells do not contain obstacles.
Input
The input consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets.
For each data set, the first line contains two positive integer numbers m and n separated by space (1m, n
20). The second line contains an integer number k(0
k
20). The ith line of the next m lines contains n integer aij separated by space (i = 1, 2,..., m;j = 1, 2,...,n). The value of aij is 1 if there is an obstacle on the cell (i, j), and is 0 otherwise.
Output
For each data set, if there exists a way for the robot to reach the cell (m, n), write in one line the integer number s, which is the number of moves the robot has to make; -1 otherwise.
题目大意：给个m*n的方格，每个格里要么是1，要么就是0, 1代表障碍物。机器人可以跨过连续的k个障碍物，求机器人从(0，0)点到(m,n)点最短的路径
解题思路：找最优解选择BFS算法，但要注意障碍物的限制。
代码：
1 #include<iostream>
2 #include<queue>
3 #include<cstring>
4 #include<string>
5 using namespace std;
6 const int maxn=20+5;
7 int bad[maxn][maxn],p[maxn][maxn][maxn];
8 int go[4][2]={0,1,1,0,-1,0,0,-1};
9 int m,n,k;
10 int check(int x,int y)
11 {
12 if(x<0 || y<0 || x>=m || y>=n)
13 return 1;
14 return 0;
15 }
16 struct node
17 {
18 int x;
19 int y;
20 int t;
21 int d;
22 }temp,temp1,temp2;
23 int main()
24 {
25 int T,dx,dy,i,j;
26 cin>>T;
27 while(T--)
28 {
29 memset(p,0,sizeof(p));
30 cin>>m>>n>>k;
31 p[0][0][0]=1;
32 for(i=0;i<m;i++)
33 for(j=0;j<n;j++)
34 cin>>bad[i][j];
35 dx=m-1;
36 dy=n-1;
37 struct node temp={0,0,0,0};
38 queue<node>q;
39 q.push(temp);
40 while(!q.empty())
41 {
42 temp1=q.front();
43 q.pop();
44 if(temp1.x==dx&&temp1.y==dy)
45 {
46 cout<<temp1.t<<endl;
47 break;
48 }
49 for(i=0;i<4;i++)
50 {
51 temp2.x=temp1.x+go[i][0];
52 temp2.y=temp1.y+go[i][1];
53 temp2.t=temp1.t+1;
54 if(bad[temp2.x][temp2.y])
55 temp2.d=temp1.d+1;
56 else
57 temp2.d=0;
58 if(check(temp2.x,temp2.y))
59 continue;
60 if(p[temp2.x][temp2.y][temp2.d]==1)
61 continue;
62 if(temp2.d<=k)
63 {
64 p[temp2.x][temp2.y][temp2.d]=1;
65 q.push(temp2);
66 }
67
68
69 }
70 if(q.empty())
71 cout<<-1<<endl;
72 }
73 }
74 return 0;
75 }