Hdu 1885 Key Task（状压+BFS）

来源：互联网发布：java实现rsa算法编辑：程序博客网时间：2024/05/18 01:40

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1885

思路：用四位二进制数每一位表示是否拥有一种钥匙，若当前状态与当前门取&不为0表示可通过；若遇到钥匙，则取|，表示拥有此种钥匙。用v[x][y][state]表示在（x，y）位置，拥有钥匙状态。

#include<map>#include<cstdio>#include<queue>#include<cstring>#include<iostream>#include<algorithm>#define debuusing namespace std;const int maxn=100+10;const int dx[]= {-1,1,0,0};const int dy[]= {0,0,-1,1};struct Node{    int x,y,state,step;    Node(int x=0,int y=0,int state=0,int step=0):x(x),y(y),state(state),step(step) {}};int n,m,stx,sty;queue<Node> q;char g[maxn][maxn];map<char,int> col,COL;int v[maxn][maxn][35];int inside(int x,int y){    return x>=1&&x<=n&&y>=1&&y<=m&&g[x][y]!='#';}int solve(int x,int y){    memset(v,0,sizeof(v));    while(!q.empty()) q.pop();    q.push(Node(x,y,0,0));    v[x][y][0]=1;    while(!q.empty())    {        Node now=q.front();        q.pop();        for(int i=0; i<4; i++)        {            int x=now.x+dx[i];            int y=now.y+dy[i];            int state=now.state;            int step=now.step+1;            if(inside(x,y))            {                //cout<<"* "<<now.x<<" "<<now.y<<" "<<state<<" "<<x<<" "<<y<<" "<<g[x][y]<<" "<<state<<endl;                switch (g[x][y])                {                    case 'B':                    case 'Y':                    case 'R':                    case 'G':                    {                          int tmp=state&(1<<COL[g[x][y]]);                          //cout<<now.x<<" "<<now.y<<" "<<state<<" "<<x<<" "<<y<<" "<<state<<" "<<g[x][y]<<endl;                          if(!v[x][y][state]&&tmp)                          {                              v[x][y][state]=1;                              q.push(Node(x,y,state,step));                          }                          break;                    }                    case 'b':                    case 'y':                    case 'r':                    case 'g':                    {                         int tmp=state|(1<<col[g[x][y]]);                         //cout<<now.x<<" "<<now.y<<" "<<state<<" "<<x<<" "<<y<<" "<<tmp<<" "<<g[x][y]<<endl;                         if(!v[x][y][tmp])                         {                             v[x][y][tmp]=1;                            q.push(Node(x,y,tmp,step));                         }                         break;                    }                    case '.':                    {                         //cout<<now.x<<" "<<now.y<<" "<<state<<" "<<x<<" "<<y<<" "<<state<<" "<<"."<<endl;                         if(!v[x][y][state])                         {                             v[x][y][state]=1;                             q.push(Node(x,y,state,step));                         }                             break;                    }                    case 'X':                    {                         return step;                    }                    case '*':                        {                            if(!v[x][y][state])                            {                                v[x][y][state]=1;                                q.push(Node(x,y,state,step));                            }                            break;                        }                }            }        }    }    return -1;}int main(){#ifdef debug     freopen("in.in","r",stdin);     //freopen("out.out","w",stdout);#endif // debug    col['b']=0,col['y']=1,col['r']=2,col['g']=3;    COL['B']=0,COL['Y']=1,COL['R']=2,COL['G']=3;    while(scanf("%d%d",&n,&m)!=EOF&&(n||m))    {        getchar();        stx=sty=-1;        for(int i=1; i<=n; i++)        {            for(int j=1; j<=m; j++)            {                char ch;                scanf("%c",&ch);                g[i][j]=ch;                if(ch=='*') stx=i,sty=j;            }            getchar();        }        //cout<<stx<<" "<<sty<<endl;        int ans=solve(stx,sty);        if(ans!=-1) printf("Escape possible in %d steps.\n",ans);        else printf("The poor student is trapped!\n");    }    return 0;}

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