hdu 5094 状压bfs+深坑

http://acm.hdu.edu.cn/showproblem.php?pid=5094

给出n*m矩阵

给出k个障碍，两坐标之间存在墙或门，门最多10种，状压可搞

给出s个钥匙位置及编号，相应的钥匙开相应的门，求从1,1到n,m的最短时间，不能到底则输出-1

这里有一个大坑：有可能同一个位置有多个门或者多个钥匙...

这么坑大丈夫？

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define clr1(x) memset(x,-1,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
typedef unsigned long long ULL;
const int modo = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int inf = 0x3fffffff;
const LL _inf = 1e18;
const int maxn = 55,maxm = 1<<12;
int n,m,p;
bool vis[maxn][maxn][maxm];
int g[maxn][maxn][maxn][maxn],key[maxn][maxn];//0up1down2left3right
int b[12];
struct node{
    int x,y,st,t;
    node(){};
    node(int xx,int yy,int _st,int tt):x(xx),y(yy),st(_st),t(tt){};
    bool operator < (const node &a)const{
        return a.t < t;
    }
};
int dx[] = {0,0,-1,1},
    dy[] = {-1,1,0,0};
bool in(int x,int y)
{
    return 1 <= x && x<=n && 1 <= y && y <= m;
}
void bfs()
{
    priority_queue<node> q;
    q.push(node(1,1,key[1][1],0));
    vis[1][1][key[1][1]] = 1;
 
    while(!q.empty()){
        node cur = q.top();
        q.pop();
        if(cur.x == n && cur.y == m){
            //cout<<cur.x<<','<<cur.y<<':';
            printf("%d\n",cur.t);
            return;
        }
        int x = cur.x,y = cur.y,t = cur.t,st = cur.st;
        //cout<<x<<'.'<<y<<':'<<t<<endl;
        for(int i = 0;i < 4;++i){
            int tx = x + dx[i],ty = y + dy[i];
            if(!in(tx,ty) || g[x][y][tx][ty] & 1 == 1)continue;
            if(g[x][y][tx][ty] && !(st & g[x][y][tx][ty]))continue;
            int _st = st | key[tx][ty];
            if(!vis[tx][ty][_st]){
                vis[tx][ty][_st] = 1;
                q.push(node(tx,ty,_st,t+1));
            }
        }
    }
    puts("-1");
}
void init()
{
    for(int i = 0;i < 12;++i)
        b[i] = 1<<i;
}
void work()
{
    clr0(vis),clr0(key);
    clr0(g);
    int k,s,x,y,q,x1,y1,x2,y2,st;
    RD(k);
    while(k--){
        RD2(x1,y1),RD3(x2,y2,st);
        g[x1][y1][x2][y2] |= b[st];
        g[x2][y2][x1][y1] |= b[st];
    }
    RD(s);
    while(s--){
        RD3(x,y,q);
        key[x][y] |= b[q];
    }
    bfs();
    return ;
}
int main()
{
    init();
    while(~RD3(n,m,p)){
        work();
    }
    return 0;
}