HDU 1429 (bfs+状态的位压缩)

hdu 1429 胜利大逃亡(续)

一共有10把钥匙，用10位的二进制反映钥匙的拥有情况

#include <iostream>
#include <cstring>
#include<cstdio>
#include <queue>
using namespace std;
char maze[21][21];
int n,m,t;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
int vis[1024][21][21];     //2^10-1 = 1023
typedef struct node{
    int x,y,key,step;
}Node;
queue<Node> q;
int bfs(int sx,int sy){
    while (!q.empty()) q.pop();
    vis[0][sx][sy]=1;
    Node s={sx,sy,0,0};
    q.push(s);
    while(!q.empty()){
        Node u = q.front();
        q.pop();
        int x,y,key,step,nx,ny,nkey,nstep,d;
        x=u.x; y=u.y; key=u.key; step=u.step;
        if(step >= t-1) return -1;
        for(d = 0;d < 4;d ++){
            nx=x+dx[d]; ny=y+dy[d]; nkey=key;
            if (!maze[nx][ny] || maze[nx][ny]==‘*‘) continue;
            if (isupper(maze[nx][ny]))
             {
                int t=maze[nx][ny]-‘A‘;
                if (!(key&(1<<t))) continue;      //钥匙不匹配
             }
             if (islower(maze[nx][ny]))
             {
                 int t=maze[nx][ny]-‘a‘;
                 nkey=key|(1<<t);                 //更新钥匙状态
             }
             if (vis[nkey][nx][ny]) continue;
             vis[nkey][nx][ny]=1;
             nstep=step+1;
             node v={nx,ny,nkey,nstep};
             if (maze[v.x][v.y]==‘^‘)
                return v.step;
             q.push(v);
         }
        }
        return -1;
}
int main()
{
    int x,y,ans,i,j;
    while(cin >> n >> m >> t){
        memset(vis,0,sizeof(vis));
        for (i=1;i<=n;i++)
            scanf("%s",maze[i]+1);
        for (i=1;i<=n;i++)
        for (j=1;j<=m;j++)
             if (maze[i][j]==‘@‘)
             {
                 x=i; y=j;
                 maze[i][j]=‘.‘;
             }
        ans = bfs(x,y);
        cout << ans << endl;
        for (i=0;i<=n+1;i++){
        for (j=0;j<=m+2;j++){
            cout << maze[i][j] << " ";
        }
        cout << endl;}

    }
    return 0;
}