poj 3083 Children of the Candy Corn

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

题意：迷宫问题，求从进入点到出口，分别按靠左边，右边和最短距离到达出口所学要的步数；

思路：用dfs求得靠左靠右走到达出口的步数，用bfs求最短最少到达出口的步数；

代码：

View Code

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <queue>
using namespace std;
int xx[6] = {0,1,0,-1,0};
int yy[6] = {0,0,-1,0,1};
int w = 0;
int h = 0;
int xs = 0;
int ys = 0;
int xe = 0;
int ye = 0;
int l = 0;
int r = 0;
int mi = 0;
char map[45][45] = {NULL};
int s[45][45] = {0};
bool visit[45][45] = {0};
int foward = 0;
int ll(int i)
{
    int te = (i+1)%4;
    if(te == 0)
    te = 4;
    return te;
}
int rr(int i)
{
    int te = (i-1)%4;
    if(te == 0)
    te = 4;
    return te;
}
void dfsl(int x,int y,int step,int fow)
{
   int te = ll(fow);
    if(x == xe && y == ye)
    {
    l = step;
    return ;
    }
   if(s[x+xx[te]][y+yy[te]])
   dfsl(x+xx[te],y+yy[te],step+1,te);
   else
    if(s[x+xx[fow]][y+yy[fow]])
   dfsl(x+xx[fow],y+yy[fow],step+1,fow);
   else
   {
       te = rr(fow);
       if(s[x+xx[te]][y+yy[te]])
       dfsl(x+xx[te],y+yy[te],step+1,te);
       else
       {
           te = (fow+2)%4;
           if(te == 0)
           te = 4;
       dfsl(x+xx[te],y+yy[te],step+1,te);
       }
   }

}
void dfsr(int x,int y,int step,int fow)
{
   int te = rr(fow);
    if(x == xe && y == ye)
    {
    r = step;
    return ;
    }
   if(s[x+xx[te]][y+yy[te]])
   dfsr(x+xx[te],y+yy[te],step+1,te);
   else
    if(s[x+xx[fow]][y+yy[fow]])
   dfsr(x+xx[fow],y+yy[fow],step+1,fow);
   else
   {
       te = ll(fow);
       if(s[x+xx[te]][y+yy[te]])
       dfsr(x+xx[te],y+yy[te],step+1,te);
       else
       {
           te = (fow+2)%4;
           if(te == 0)
           te = 4;
       dfsr(x+xx[te],y+yy[te],step+1,te);
       }
   }

}
void bfs(int x,int y,int step,int fow)
{
    queue<int>x3;
    queue<int>y3;
    queue<int>ste;
    queue<int>fo;
    x3.push(x);
    y3.push(y);
    ste.push(step);
    fo.push(fow);
    visit[x][y] = 1;
    while(!x3.empty())
    {
        x = x3.front();
        y = y3.front();
        step = ste.front();
        fow = fo.front();
        x3.pop();
       y3.pop();
        ste.pop();
        fo.pop();
        if(x == xe&&y == ye)
        {
            mi =step;
        return ;
        }
        mi = step;
        int te;
        te = rr(fow);
        if(!visit[x+xx[te]][y+yy[te]]&&s[x+xx[te]][y+yy[te]])
        {
            visit[x+xx[te]][y+yy[te]] = 1;
        x3.push(x+xx[te]);
        y3.push(y+yy[te]);
        ste.push(step+1);
        fo.push(te);
        }
        if(!visit[x+xx[fow]][y+yy[fow]]&&s[x+xx[fow]][y+yy[fow]])
        {
            visit[x+xx[fow]][y+yy[fow]] = 1;
          x3.push(x+xx[fow]);
        y3.push(y+yy[fow]);
        ste.push(step+1);
        fo.push(fow);
        }
        te = ll(fow);
        if(!visit[x+xx[te]][y+yy[te]]&&s[x+xx[te]][y+yy[te]])
        {
            visit[x+xx[te]][y+yy[te]] = 1;
        x3.push(x+xx[te]);
        y3.push(y+yy[te]);
        ste.push(step+1);
        fo.push(te);
        }
    }
}
int main()
{
    int num = 0;
    scanf("%d",&num);
    while(num--)
    {
        memset(s,0,sizeof(s));
        memset(visit,0,sizeof(visit));
        scanf("%d%d",&w,&h);
        getchar();
        for(int i = h-1;i >= 0; --i)
        {
           scanf("%s",map[i]);
        }
        for(int i = 0;i < h; ++i)
        for(int j = 0;j < w; ++j)
        if(map[i][j] != '#')
        {
        s[i][j] = 1;
        if(map[i][j] == 'S')
        {
            xs = i;
            ys = j;
        }
        if(map[i][j] == 'E')
        {
            xe = i;
            ye = j;
        }
        }
        if(xs == 0)
        foward = 1;
        if(xs == h-1)
        foward = 3;
        if(ys == 0)
        foward = 4;
        if(ys == w-1)
        foward = 2;
        dfsl(xs,ys,1,foward);
        dfsr(xs,ys,1,foward);
        bfs(xs,ys,1,foward);
        printf("%d %d %d\n",l,r,mi);
    }
    return 0;
}