hdu5094Maze bfs+状态压缩

//开一个四维数组记录墙和门的情况

//开一个三维数组标记在该位置时有哪些钥匙

//钥匙的记录用状态压缩

//注意在同一个位置可以有多把钥匙,在这卡了一个晚上。。。。。。。

#include<iostream>

#include<cstdio>

#include<cstring>

#include<queue>

using namespace std;

const int maxn = 60 ;

int vis[maxn][maxn][1<<11] ;

int wall[maxn][maxn][maxn][maxn] ;

int maze[maxn][maxn] ;

struct node

{

int x , y;

int step ;

int state ;

};

int dx[4] = {-1,0,0,1} ;

int dy[4] = {0,1,-1,0} ;

int n ,m,p;

int bfs()

{

queue<struct node> que ;

memset(vis,0,sizeof(vis));

struct node first = {1,1,0,0};

vis[1][1][0] = 1;

que.push(first) ;

while(que.size())

{

struct node now = que.front() ;

que.pop() ;

if(now.x == n && now.y == m)

return now.step ;

for(int i = 0;i < 4;i++)

{

struct node next ;

next.x = now.x +dx[i];

next.y = now.y +dy[i];

if(next.x<1 || next.x>n || next.y>m || next.y<1 || wall[now.x][now.y][next.x][next.y] == 0)

continue ;

if(wall[now.x][now.y][next.x][next.y] >= 1)

if(!(now.state & (1<<(wall[now.x][now.y][next.x][next.y]))))

continue;

next.state = now.state ;

if(maze[next.x][next.y] > 0)

next.state |= maze[next.x][next.y];

next.step = now.step + 1;

if(vis[next.x][next.y][next.state])

continue ;

vis[next.x][next.y][next.state] = 1;

que.push(next) ;

}

}

return -1;

}

int main()

{

//freopen("in.txt","r",stdin);

//  freopen("out.txt","w",stdout);

while(scanf("%d%d%d", &n , &m , &p) != EOF)

{

int k ;

scanf("%d", &k);

int xi1 , yi1, xi2 , yi2 , gi;

memset(wall,-1,sizeof(wall)) ;

memset(maze,0,sizeof(maze)) ;

for(int i = 1;i <= k ;i++)

{

scanf("%d%d%d%d%d",&xi1,&yi1,&xi2,&yi2,&gi);

wall[xi1][yi1][xi2][yi2] = gi ;

wall[xi2][yi2][xi1][yi1] = gi;

}

int s;

scanf("%d", &s) ;

int xi,yi;

for(int i = 1;i <= s;i++)

{

scanf("%d%d%d",&xi,&yi,&gi) ;

maze[xi][yi] |= (1<<(gi)) ;

}

printf("%d\n",bfs()) ;

}

}