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;

}