POJ3322 Bloxorz I 无脑广搜（我死了。。。）

多测不清空，爆零两行泪。。。。我死了QWQ

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每个节点3个状态：横坐标，纵坐标，和方向

　　说一下方向：0:立着，1:竖着躺着，上半部分在(x,y)，2:横着躺着，左半部分在(x,y)

然后就有了常量数组：

const int dx[][]={{-,,,},{-,,,},{-,,,}};
const int dy[][]={{,-,,},{,-,,},{,-,,}};
const int dz[][]={{,,,},{,,,},{,,,}};

第一维是状态中的方向，第二维是要扩展的方向（0，1，2，3）

然后就搜他。。。。记得queue要清零，d要清零，sz要清零。。。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define R register int
const int dx[][]={{-,,,},{-,,,},{-,,,}};
const int dy[][]={{,-,,},{,-,,},{,-,,}};
const int dz[][]={{,,,},{,,,},{,,,}};
using namespace std;
inline int g() {
    R ret=; register char ch; while(!isdigit(ch=getchar()));
    do ret=ret*+(ch^); while(isdigit(ch=getchar())); return ret;
}
struct node{
    int x,y,z; node() {}
    node(int xx,int yy,int zz):x(xx),y(yy),z(zz) {}
};
queue<node>q;
char e[][];
int n,m,sx,sy,sz,ex,ey;
int d[][][];
inline bool ckpos(int x,int y) {return x>&&x<=n&&y>&&y<=m;}
inline bool ck(int x,int y,int z) {
    if(!ckpos(x,y)||e[x][y]=='#'||d[x][y][z]!=-) return false;
    if(z==&&e[x][y]=='E') return false;
    if(z==&&(!ckpos(x+,y)||e[x+][y]=='#')) return false;
    if(z==&&(!ckpos(x,y+)||e[x][y+]=='#')) return false; return true;
}
int bfs() {
    memset(d,-,sizeof(d)); while(q.size()) q.pop();
    q.push(node(sx,sy,sz)); d[sx][sy][sz]=;
    while(q.size()) { node u=q.front(); q.pop();
        for(R i=;i<;++i) {
            node v=node(u.x+dx[u.z][i],u.y+dy[u.z][i],dz[u.z][i]); //cout<<u.x<<" "<<u.y<<" "<<u.z<<"  "<<v.x<<" "<<v.y<<" "<<v.z<<endl;
            if(!ck(v.x,v.y,v.z)) continue;
            q.push(v),d[v.x][v.y][v.z]=d[u.x][u.y][u.z]+;
            if(v.x==ex&&v.y==ey&&v.z==) return d[v.x][v.y][v.z];
        }
    } return -;
}
signed main() {
    while(n=g(),m=g(),n!=) { sz=;
        for(R i=;i<=n;++i) scanf("%s",e[i]+);
        for(R i=;i<=n;++i) for(R j=;j<=m;++j)
            if(e[i][j]=='X') { sx=i,sy=j; e[i][j]='.';
                if(j<m&&e[i][j+]=='X') sz=,e[i][j+]='.';
                if(i<n&&e[i+][j]=='X') sz=,e[i+][j]='.';
            } else if(e[i][j]=='O') ex=i,ey=j,e[i][j]='.';
        R ans=bfs(); ans==-?printf("Impossible\n"):printf("%d\n",ans);
        //for(R i=1;i<=n;++i,cout<<'\n') for(R j=1;j<=m;++j) cout<<d[i][j][0]<<" "<<d[i][j][1]<<" "<<d[i][j][2]<<" ";
    }
}

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2019.04.26