HDU1429 bfs 状态压缩 xingxing在努力

　　这道题在传统的走迷宫问题上面加了门和钥匙， 那我们在加一个状态表示他目前所携带的钥匙即可。。代码如下：

　　wa点：1.在用钥匙打开门后钥匙还是可以继续带在身上的 2：判断一个状态某一位我1是（st>>num）&1 == 1..不要搞错了啊

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>

using namespace std;
int n, m, t;
char map[50][50];
int sx, sy, ex, ey;
int vis[21][21][1027];
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
bool inside(int x, int y) { return x>=0&&x<n&&y>=0&&y<m; }


struct state
{
    int x, y, steps;
    int st;
    state() {}
    state(int x, int y, int steps, int st):x(x), y(y), steps(steps), st(st) {}
};

void debug(state tp)
{
    printf("(x, y)=(%d, %d), steps=%d, state=%d
", tp.x, tp.y, tp.steps, tp.st);
}

queue<state> que;

int bfs()
{
    memset(vis, 0, sizeof(vis));
    while(!que.empty())     que.pop();
    que.push(state(sx, sy, 0, 0));
    vis[sx][sy][0] = 1;
    while(!que.empty())
    {
        state tp = que.front(); que.pop();
        //printf("POP: ");
        //debug(tp);
        if(tp.x==ex && tp.y==ey)
            return tp.steps;
        /*if(tp.x==1&&tp.y==3&&tp.steps==6&&tp.st==2)
        {
            int tp;
            tp = 2;
        }*/
        for(int i=0; i<4; i++)
        {
            int nx = tp.x + dx[i], ny = tp.y + dy[i], steps = tp.steps;
            int nowstate = tp.st;
            if(!inside(nx, ny)) 
                continue;
            if(map[nx][ny]=='.' && !vis[nx][ny][nowstate])
            {
                que.push(state(nx, ny, steps+1, nowstate));
                vis[nx][ny][nowstate] = 1;
                //printf("  push: ");
                //debug(state(nx, ny, steps+1, nowstate));
            }
            else if(map[nx][ny] == '*')             //遇到了墙
                continue;
            else if(map[nx][ny]>='A' && map[nx][ny]<='J')   //遇到了门
            {
                int num = map[nx][ny] - 'A';
                if(((nowstate>>num)&1) == 1 && !vis[nx][ny][nowstate])   //可以打开
                {
                    que.push(state(nx, ny, steps+1, nowstate));
                    vis[nx][ny][nowstate] = 1;
                    //printf("  push: ");
                    //debug(state(nx, ny, steps+1, nowstate^(1<<num)));
                }
            }
            else if(map[nx][ny]>='a' && map[nx][ny]<='j')
            {
                int num = map[nx][ny] - 'a';
                if(!vis[nx][ny][nowstate|(1<<num)])
                {
                    que.push(state(nx, ny, steps+1, nowstate|(1<<num)));
                    vis[nx][ny][nowstate|(1<<num)] = 1;
                    //printf("  push: ");
                    //debug(state(nx, ny, steps+1, nowstate|(1<<num)));
                }
            }
        }
    }    
    return -1;
}

int main()
{    
    
    while(scanf("%d%d%d", &n, &m, &t) == 3)
    {    
        for(int i=0; i<n; i++)
        {
            scanf("%s", map[i]);
            for(int j=0; j<m; j++)
                if(map[i][j] == '@')
                {
                    sx=i, sy=j;
                    map[i][j] = '.';
                }
                else if(map[i][j] == '^')
                {
                    ex=i, ey=j;
                    map[i][j] = '.';
                }
        }
        int num = bfs();
        if(num>=0 && num<t)
            printf("%d
", num);
        else 
            printf("-1
");
    }    
    return 0;
}