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\n", 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\n", num);
else
printf("-1\n");
}
return 0;
}