hdu 5094 状压bfs+深坑

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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-9const 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];//0up1down2left3rightint 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;}

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