Hdu1429 victory (continued) (State compression + BFS)

Problem DescriptionIgnatius is again taken away by the Devil (I don't know why he is so fond of the devil )......

This time, the devil learned the last lesson, locked Ignatius in an n * m dungeon, installed locks in some places in the dungeon, and hid the keys in some other places in the dungeon. At first Ignatius was shut down in the position of (sx, sy) and left the warehouse's door in the position of (ex, ey. Each minute, Ignatius can only go from one coordinate to one of the four adjacent coordinates. The magic Lord visits the dungeon once every t minutes. If it finds that Ignatius is not in the original position, he will be taken back. After several attempts, Ignatius has drawn a map of the entire dungeon. Please help him calculate whether he can successfully escape again. As long as the devil leaves the dungeon before the next visit, it will be deemed as a failed escape if the devil just leaves the exit or has not yet reached the exit.

The first row of test data in each Input group has three integers: n, m, t (2 <= n, m <= 20, t> 0 ). The next n rows of m column are the map of the dungeon, including:

. Represents the path
* Representing the wall
@ Indicates the starting position of Ignatius
^ Represents the exit of the dungeon
The A-J represents the door with a lock, and the corresponding keys are a-j
A-j Represents the key, and the corresponding door is the A-J

There is a blank line between each group of test data.

Output for each group of test data. If the data can be successfully escaped, Output how many minutes to exit. If not, Output-1.

Sample Input

4 5 17@A.B.a*.*.*..*^c..b*4 5 16@A.B.a*.*.*..*^c..b*

Sample Output

16-1 each location has different key statuses.
#include
  
   #include
   
    #include
    
     #include
     
      using namespace std;typedef struct nnn{    int t,key,x,y;}NODE;int Key[25][25][1<<10],T,n,m;char map[25][25];int abs(int x){    return x>=0?x:-x;}int bfs(int sx,int sy){    queue
      
       q;    NODE now,pre;    int dir[4][2]={0,1,0,-1,1,0,-1,0};    int tx,ty;    now.key=0; now.t=0; now.x=sx; now.y=sy;    Key[sy][sx][0]=1;    q.push(now);    while(!q.empty())    {        pre=q.front(); q.pop();        for(int i=0; i<4; i++)        {            now=pre;            ty=pre.y+dir[i][0]; tx=pre.x+dir[i][1];            if(ty>=0&&ty
       
        =0&&tx
        
         ='a'&&map[ty][tx]<='j'){ now.key|=(1<<(map[ty][tx]-'a')); if(Key[ty][tx][now.key]==0) Key[ty][tx][now.key]=1,q.push(now); } else if(map[ty][tx]>='A'&&map[ty][tx]<='J') { if(Key[ty][tx][now.key]==0&&(pre.key&(1<<(map[ty][tx]-'A')))) Key[ty][tx][now.key]=1, q.push(now); } else if(Key[ty][tx][now.key]==0){ Key[ty][tx][now.key]=1; q.push(now); } } } } return -1;}int main(){ int sx,sy,ex,ey; while(scanf("%d%d%d",&n,&m,&T)>0) { for(int i=0; i
         
          =T) { printf("-1\n"); continue; } memset(Key,0,sizeof(Key)); printf("%d\n",bfs(sx,sy)); }}