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[3][4]={{-2,0,1,0},{-1,0,2,0},{-1,0,1,0}};
const int dy[3][4]={{0,-2,0,1},{0,-1,0,1},{0,-1,0,2}};
const int dz[3][4]={{1,2,1,2},{0,1,0,1},{2,0,2,0}};

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[3][4]={{-2,0,1,0},{-1,0,2,0},{-1,0,1,0}};
const int dy[3][4]={{0,-2,0,1},{0,-1,0,1},{0,-1,0,2}};
const int dz[3][4]={{1,2,1,2},{0,1,0,1},{2,0,2,0}};
using namespace std;
inline int g() {
    R ret=0; register char ch; while(!isdigit(ch=getchar()));
    do ret=ret*10+(ch^48); 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[510][510];
int n,m,sx,sy,sz,ex,ey;
int d[510][510][3];
inline bool ckpos(int x,int y) {return x>0&&x<=n&&y>0&&y<=m;}
inline bool ck(int x,int y,int z) {
    if(!ckpos(x,y)||e[x][y]=='#'||d[x][y][z]!=-1) return false;
    if(z==0&&e[x][y]=='E') return false;
    if(z==1&&(!ckpos(x+1,y)||e[x+1][y]=='#')) return false;
    if(z==2&&(!ckpos(x,y+1)||e[x][y+1]=='#')) return false; return true;
}
int bfs() {
    memset(d,-1,sizeof(d)); while(q.size()) q.pop();
    q.push(node(sx,sy,sz)); d[sx][sy][sz]=0;
    while(q.size()) { node u=q.front(); q.pop();
        for(R i=0;i<4;++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]+1;
            if(v.x==ex&&v.y==ey&&v.z==0) return d[v.x][v.y][v.z];
        }
    } return -1;
}
signed main() {
    while(n=g(),m=g(),n!=0) { sz=0;
        for(R i=1;i<=n;++i) scanf("%s",e[i]+1);
        for(R i=1;i<=n;++i) for(R j=1;j<=m;++j) 
            if(e[i][j]=='X') { sx=i,sy=j; e[i][j]='.';
                if(j<m&&e[i][j+1]=='X') sz=2,e[i][j+1]='.';
                if(i<n&&e[i+1][j]=='X') sz=1,e[i+1][j]='.'; 
            } else if(e[i][j]=='O') ex=i,ey=j,e[i][j]='.';
        R ans=bfs(); ans==-1?printf("Impossible
"):printf("%d
",ans);
        //for(R i=1;i<=n;++i,cout<<'
') 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