HDU 1885 Key Task

这题就是找到对应钥匙然后突破对应的门，反正最终目的是要到达X 这个终点。要求花费最小时间，如果不能到达输出The poor student is trapped!

很明显是广搜，把钥匙的四种状态都加入到哈希表来记重，还有对应的坐标所以是六维数组。。

#include<algorithm>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
int n, m;
char map[105][105];
bool vis[105][105][2][2][2][2];
struct node
{
    node(int p,int q,bool a1,bool a2,bool a3,bool a4,int a5)
    {
        x=p, y=q, b=a1, ye=a2, r=a3, g=a4, now=a5;
    }
    node() {}
    int x, y, now;
    bool b, ye, r, g;

};
bool can(int x,int y)
{
    if(x<0||y<0||x>=n||y>=m)
        return false;
    if(map[x][y]=='#')
        return false;
    return true;
}
int dx[]= {1,0,-1,0}, dy[]= {0,1,0,-1};
int main()
{
    int i, j, sx, sy;
    bool flag;
    while(~scanf("%d%d",&n,&m),n&&m)
    {
        queue<node> q;
        flag=false;
        for(i=0; i<n; i++)
        {
            scanf("%s",map[i]);
            for(j=0; j<m; j++)
            {
                if(map[i][j]=='*')
                    sx=i, sy=j;
                if(map[i][j]=='X')
                    flag=true;
            }
        }
        if(!flag)
        {
            puts("The poor student is trapped!");
            continue;
        }
        memset(vis,0,sizeof(vis));
        vis[sx][sy][0][0][0][0]=true;
        q.push(node(sx,sy,false,false,false,false,0));
        node tmp;
        map[sx][sy]='.';
        flag=false;
        int x, y, nx, ny, now, cc;
        bool b, ye, r, g;
        while(!q.empty())
        {
            tmp=q.front();
            q.pop();
            x=tmp.x, y=tmp.y, b=tmp.b, ye=tmp.ye, r=tmp.r, g=tmp.g, now=tmp.now;
            bool vb, vye, vr, vg;
            if(map[x][y]=='X')
            {
                flag=true;
                cc=now;
                break;
            }
            for(i=0; i<4; i++)
            {
                nx=x+dx[i], ny=y+dy[i];
                if(!can(nx,ny)) continue;
                if(map[nx][ny]=='.'||map[nx][ny]=='X')
                {
                    if(!vis[nx][ny][b][ye][r][g])
                        q.push(node(nx,ny,b,ye,r,g,now+1));
                    vis[nx][ny][b][ye][r][g]=true;
                }
                else if(map[nx][ny]>='a'&&map[nx][ny]<='z')
                {
                    vb=b, vye=ye, vr=r, vg=g;
                    if(map[nx][ny]=='b')
                        vb=true;
                    if(map[nx][ny]=='y')
                        vye=true;
                    if(map[nx][ny]=='r')
                        vr=true;
                    if(map[nx][ny]=='g')
                        vg=true;
                    if(!vis[nx][ny][b][ye][r][g])
                        q.push(node(nx,ny,vb,vye,vr,vg,now+1));
                    vis[nx][ny][b][ye][r][g]=true;
                }
                else if(map[nx][ny]>='A'&&map[nx][ny]<='Z')
                {
                    bool ans=false;
                    char kk=map[nx][ny];
                    if(kk=='B'&&b)
                        ans=true;
                    if(kk=='Y'&&ye)
                        ans=true;
                    if(kk=='R'&&r)
                        ans=true;
                    if(kk=='G'&&g)
                        ans=true;
                    if(ans)
                    {
                        if(!vis[nx][ny][b][ye][r][g])
                            q.push(node(nx,ny,b,ye,r,g,now+1));
                        vis[nx][ny][b][ye][r][g]=true;
                    }
                }
            }
        }
        if(flag) printf("Escape possible in %d steps.\n",cc);
        else puts("The poor student is trapped!");
    }
    return 0;
}