HDU 1180(简单BFS）

题目意思有点不清 其实是可以在原地停留的 注意桥的变化与时间相关

 

用方向数组压缩代码  OK 以下是实现

 

 

 

#include <iostream>
#include <queue>
using namespace std;

char Map[21][21];
typedef struct  
{
    long mi,mj;
    long step;
}Node;

Node Start,End;

bool hash[21][21];

long dx[4]={0,0,1,-1};
long dy[4]={1,-1,0,0};

long M,N;

inline void BFS()
{
    memset(hash,0,sizeof(hash));

    queue<Node> q;

    q.push(Start);

    Node t;

    long cost=0;

    hash[Start.mi][Start.mj]=true;

    while (!q.empty())
    {
        t=q.front();
        q.pop();

        if (End.mi==t.mi&&End.mj==t.mj)
        {
            cost=t.step;
            break;
        }

        long i;
        for (i=0;i<4;++i)
        {
            Node n;
            n.mi=t.mi+dx[i];
            n.mj=t.mj+dy[i];

            if (n.mi>=0&&n.mi<M&&n.mj>=0&&n.mj<N&&Map[n.mi][n.mj]!='*'&&!hash[n.mi][n.mj])
            {
                if (Map[n.mi][n.mj]=='.'||Map[n.mi][n.mj]=='T')
                {
                    n.step=t.step+1;
                    q.push(n);    
                    hash[n.mi][n.mj]=true;
                }

                if (Map[n.mi][n.mj]=='-')
                {
                    if ((t.step&1)==0)
                    {
                        if(n.mi==t.mi)
                        {
                            n.mj=n.mj+n.mj-t.mj;
                            n.step=t.step+1;
                            
                            if(n.mi>=0&&n.mi<M&&n.mj>=0&&n.mj<N&&!hash[n.mi][n.mj]&&Map[n.mi][n.mj]!='*')
                            {
                                q.push(n);
                                hash[n.mi][n.mj]=true;
                                hash[t.mi][t.mj]=true;
                            }
                        }
                        else
                        {
                            n.mi=t.mi;
                            n.mj=t.mj;
                            n.step=t.step+1;
                            q.push(n);
                        }
                    }
                    else
                    {
                        if(n.mj==t.mj)
                        {
                            n.mi=n.mi+n.mi-t.mi;
                            n.step=t.step+1;        
                            if(n.mi>=0&&n.mi<M&&n.mj>=0&&n.mj<N&&!hash[n.mi][n.mj]&&Map[n.mi][n.mj]!='*')
                            {
                                q.push(n);
                                hash[n.mi][n.mj]=true;
                                hash[t.mi][t.mj]=true;
                            }
                        }
                        else
                        {
                            n.mi=t.mi;
                            n.mj=t.mj;
                            n.step=t.step+1;
                            q.push(n);
                        }
                    }
                }

                if (Map[n.mi][n.mj]=='|')
                {
                    if (t.step&1)
                    {
                        if(n.mi==t.mi)
                        {
                            n.mj=n.mj+n.mj-t.mj;
                            n.step=t.step+1;
                            
                            if(n.mi>=0&&n.mi<M&&n.mj>=0&&n.mj<N&&!hash[n.mi][n.mj]&&Map[n.mi][n.mj]!='*')
                            {
                                q.push(n);
                                hash[n.mi][n.mj]=true;
                                hash[t.mi][t.mj]=true;
                            }
                        }
                        else
                        {
                            n.mi=t.mi;
                            n.mj=t.mj;
                            n.step=t.step+1;
                            q.push(n);
                        }
                    }
                    else
                    {
                        if(n.mj==t.mj)
                        {
                            n.mi=n.mi+n.mi-t.mi;
                            n.step=t.step+1;
                            
                            if(n.mi>=0&&n.mi<M&&n.mj>=0&&n.mj<N&&!hash[n.mi][n.mj]&&Map[n.mi][n.mj]!='*')
                            {
                                q.push(n);
                                hash[n.mi][n.mj]=true;
                                hash[t.mi][t.mj]=true;
                            }
                        }
                        else
                        {
                            n.mi=t.mi;
                            n.mj=t.mj;
                            n.step=t.step+1;
                            q.push(n);
                        }
                    }

                }
            }
        }    
    }
    //题目保证可达
    printf("%ld\n",cost);

}

int main()
{
    
    while (scanf("%ld %ld",&M,&N)!=EOF)
    {
        long i,j;
        getchar();
        for (i=0;i<M;++i)
        {
            gets(Map[i]);
            for (j=0;j<N;++j)
            {
                if (Map[i][j]=='S')
                {
                    Start.mi=i;
                    Start.mj=j;
                    Start.step=0;
                }

                if (Map[i][j]=='T')
                {
                    End.mi=i;
                    End.mj=j;
                }
            }
        }

        BFS();

    }

    return 0;
}