hdu1429胜利大逃亡(续) (状态压缩+BFS)

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

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

16
-1
每个位置都有不同钥匙的状态经过。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
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<NODE>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<n&&tx>=0&&tx<m&&pre.t<T-1&&map[ty][tx]!=‘*‘)
            {
                now.t++; now.y=ty; now.x=tx;
                if(map[ty][tx]==‘^‘) return now.t;
                if(map[ty][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<n; i++)
        scanf("%s",map[i]);
        for(int y=0 ; y<n; y++)
        for(int x=0 ; x<m; x++)
        if(map[y][x]==‘@‘)
        sy=y,sx=x;
        else if(map[y][x]==‘^‘)
        ey=y,ex=x;
        if(abs(sy-ey)+abs(sx-ex)>=T)
        {
            printf("-1\n"); continue;
        }
        memset(Key,0,sizeof(Key));
        printf("%d\n",bfs(sx,sy));
    }
}