紫书UVA1600

这个题的第二组数据一直过不了，原因是int layer=t.layer写在了for的外面，导致每一个方向共用了一个layer。

/*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
(1 ≤ m, n ≤ 20). The second line contains an integer number k (0 ≤ k ≤ 20). The i-th 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.
Sample Input
3
2 5
0
0 1 0 0 0
0 0 0 1 0
4 6
1
0 1 1 0 0 0
0 0 1 0 1 1
0 1 1 1 1 0
0 1 1 1 0 0
2 2
0
0 1
1 0
Sample Output
7
10
-1*/
#include <bits/stdc++.h>

using namespace std;
int a[25][25],vis[25][25][25];
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
int m,n,k;
struct node
{
    int x,y,layer,step;
    node(int x=1,int y=1,int l=0,int s=0):x(x),y(y),layer(l),step(s){}
};
int cas=0;
int bfs()
{
    queue <node> q;
    memset(vis,0,sizeof(vis));
    q.push(node(1,1,0,0));
   // vis[1][1][0]=1;
    while(!q.empty())
    {
        node t=q.front();
        int x=t.x;
        int y=t.y;

        int step=t.step;
        q.pop();
        if(x==m&&y==n)
            return step;
           // cout<<t.step<<endl;
        for(int i=0;i<4;i++)
        {
            int x1=x+dx[i];
            int y1=y+dy[i];
            int layer=t.layer;//此处没写对位置导致我第二组数据答案出错
            if(a[x1][y1])
                layer++;
            else
                layer=0;
            if(x1>0&&x1<=m&&y1>0&&y1<=n&&!vis[x1][y1][layer]&&layer<=k)
            {
                vis[x1][y1][layer]=1;
                q.push(node(x1,y1,layer,step+1));
            }
        }
    }
    return -1;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>m>>n>>k;
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                cin>>a[i][j];
            }
        }
        cout<<bfs()<<endl;
    }
    return 0;

}