poj 3083 Children of the Candy Corn

http://poj.org/problem?id=3083

题目不难 就是繁琐呀

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<string>
#include<queue>
#include<stack>
#include <iomanip>
using namespace std;
#define LL long long
#define sint short int
const int INF=0x3f3f3f3f;
//priority_queue<int,vector<int>,greater<int> >qt;
const int N=50;
char graph[N][N];
typedef pair<int,int>point;
queue<point>qt;
int dist[N][N];
bool in[N][N];
int X[]={-1,0,1,0};
int Y[]={0,1,0,-1};
int n,m;
bool OK(int x,int y)
{
    if(x>=0&&x<n&&y>=0&&y<m&&graph[x][y]!='#')
    return true;
    return false;
}
int spfa(int sx,int sy,int ex,int ey)
{
    memset(in,false,sizeof(in));
    memset(dist,-1,sizeof(dist));
    qt.push(point(sx,sy));
    in[sx][sy]=true;
    dist[sx][sy]=1;
    while(!qt.empty())
    {
        int x=qt.front().first;
        int y=qt.front().second;
        in[x][y]=false;
        qt.pop();
        for(int i=0;i<4;++i)
        {
            int l1=x+X[i];
            int l2=y+Y[i];
            if(OK(l1,l2)&&(dist[l1][l2]==-1||dist[l1][l2]>dist[x][y]+1))
            {
               dist[l1][l2]=dist[x][y]+1;
               if(!in[l1][l2])
               {
                   in[l1][l2]=true;
                   qt.push(point(l1,l2));
               }
            }
        }
    }
    return dist[ex][ey];
}
int Lsearch(int sx,int sy,int ex,int ey,int t)
{
    int step=1;
    int kx=sx,ky=sy;
    while(true)
    {
        if(kx==ex&&ky==ey)
        break;
        bool flag=false;
        for(int i=(t+3)%4,j=0;j<4;++j,i++)
        {
            int l=(i+4)%4;
            int l1=kx+X[l];
            int l2=ky+Y[l];
            if(OK(l1,l2))
            {kx=l1;ky=l2;t=l;++step;flag=true;break;}
        }
        if(flag==false)
        return INF;
    }
    return step;
}
int Rsearch(int sx,int sy,int ex,int ey,int t)
{//cout<<kx<<" "<<ky<<" "<<t<<endl;
    int step=1;
    int kx=sx,ky=sy;
    while(true)
    {//cout<<kx<<" "<<ky<<" "<<t<<endl;
        if(kx==ex&&ky==ey)
        break;
        bool flag=false;
        for(int i=(t+1)%4,j=0;j<4;++j,i--)
        {
            int l=(i+4)%4;
            int l1=kx+X[l];
            int l2=ky+Y[l];
            if(OK(l1,l2))
            {kx=l1;ky=l2;t=l;++step;flag=true;break;}
        }
        if(flag==false)
        return INF;
    }
    return step;
}
int main()
{
    //freopen("data.in","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int sx,sy,ex,ey;
        scanf("%d %d",&m,&n);
        getchar();
        for(int i=0;i<n;++i)
        gets(graph[i]);
        for(int i=0;i<n;++i)
        for(int j=0;j<m;++j)
        if(graph[i][j]=='S')
        {sx=i;sy=j;}
        else if(graph[i][j]=='E')
        {ex=i;ey=j;}
        int ans1=INF,ans2=INF,ans3;
        //cout<<sx<<" "<<sy<<" "<<ex<<" "<<ey<<endl;
        for(int i=0;i<4;++i)
        {
            ans1=min(ans1,Lsearch(sx,sy,ex,ey,i));
            ans2=min(ans2,Rsearch(sx,sy,ex,ey,i));
        }
        ans3=spfa(sx,sy,ex,ey);
        printf("%d %d %d\n",ans1,ans2,ans3);
    }
    return 0;
}