poj 3083 Children of the Candy Corn

http://poj.org/problem?id=3083
题目不难 主要是繁琐
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<string>
#include<queue>
#include<algorithm>
#include<map>

using namespace std;
const int N=50;
int x[4]={0,-1,0,1};
int y[4]={-1,0,1,0};
int LL[4]={-1,0,1,2};
int RR[4]={1,0,-1,2};
int stx,sty,ndx,ndy;
bool field[N][N];
bool had[N][N];
int ans1,ans2,ans3;
int n,m;
char ctemp;
void bfs()
{
    memset(had,false,sizeof(had));
    queue<int>strx;
    queue<int>stry;
    queue<int>step;
    strx.push(stx);
    stry.push(sty);
    step.push(1);
    had[stx][sty]=true;
    while(!strx.empty())
    {
        int x1,y1,step1,l1,l2;
        x1=strx.front();
        y1=stry.front();
        step1=step.front();
        strx.pop();stry.pop();step.pop();
        for(int i=0;i<4;++i)
        {
            //cout<<x1<<" "<<y1<<endl;
           l1=x1+x[i];l2=y1+y[i];
           //cout<<l1<<" "<<l2<<endl;
           if(l1==ndx&&l2==ndy)
           {
               ans3=step1+1;
               return;
           }
           if(!had[l1][l2]&&field[l1][l2])
           {
               had[l1][l2]=true;
               //cout<<l1<<" "<<l2<<" "<<(step1+1)<<endl;
               strx.push(l1);
               stry.push(l2);
               step.push(step1+1);
           }
        }
    }
}
bool ok(int i,int j,int k1,int &i1,int &j1)
{
   i1=i+x[k1];
   j1=j+y[k1];

   if(field[i1][j1])
   return true;
   return false;
}
void dfs(char LR,int i,int j,int k,int stepn)
{
    int i1,j1,k1;
    k=k+4;
        for(int l=0;l<4;++l)
        {
            if(LR=='L')
            k1=(k+LL[l])%4;
            else
            k1=(k+RR[l])%4;
            //cout<<k<<" "<<k1<<endl;
            if(ok(i,j,k1,i1,j1))
            {//cout<<i<<" "<<j<<" "<<i1<<" "<<j1<<endl;
              if(i1==ndx&&j1==ndy)
              {
                 if(LR=='L')
                 ans1=stepn+1;
                 else
                 ans2=stepn+1;
                 return;
              }
              dfs(LR,i1,j1,k1,stepn+1);
              return;
           }
        }
    return ;
}
int main()
{
    int T;
    cin>>T;
    while(cin>>n>>m)
    {
        memset(field,false,sizeof(field));
        for(int i=1;i<=m;++i)
        {
            for(int j=1;j<=n;++j)
            {
                cin>>ctemp;
                if(ctemp!='#')
                {
                   field[i][j]=true;
                }
                if(ctemp=='S')
                {
                    stx=i;
                    sty=j;
                }else
                if(ctemp=='E')
                {
                    ndx=i;
                    ndy=j;
                }
            }
        }

            int k;
            if(field[stx+1][sty])
            {
                k=3;
            }else
            if(field[stx][sty+1])
            {
                k=2;
            }else
            if(field[stx-1][sty])
            {
                k=1;
            }else
            {
                k=0;
            }
            dfs('L',stx,sty,k,1);
            dfs('R',stx,sty,k,1);
            bfs();
            cout<<ans1<<" "<<ans2<<" "<<ans3<<endl;
    }
    return 0;
}