【搜索】 HDU 3137 No Left Turns

HDU 3137 No Left Turns"/>

【搜索】 HDU 3137 No Left Turns

题意：S到F 只能直走和右拐

求最少步数

开始可以任意选择一个方向

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <map>
#define cler(arr, val)    memset(arr, val, sizeof(arr))
typedef long long  LL;
const int MAXN = 543;
const int MAXM = 260110;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
char mp[21][21];
bool vis[4][21][21];
int n,m;
struct node
{int fx;int x,y,step;
};
queue<node>q;
bool ok(int x,int y)
{if(x>=0&&x<n&&y>=0&&y<m&&mp[x][y]!='X') return true;return false;
}
int bfs(int x,int y)
{cler(vis,false);while(!q.empty()) q.pop();node front,rear;front.x=x,front.y=y;front.step=0;for(int i=0;i<4;i++){front.fx=i;q.push(front);vis[i][x][y]=true;}while(!q.empty()){front=q.front();q.pop();int dx1=front.x,dy1=front.y,fx;int dx2=front.x,dy2=front.y;if(front.fx==0) dx1-=1,dy2+=1,fx=3;//上if(front.fx==1) dx1+=1,dy2-=1,fx=2;//下if(front.fx==2) dy1-=1,dx2-=1,fx=0;//左if(front.fx==3) dy1+=1,dx2+=1,fx=1;//右if(ok(dx1,dy1)&&!vis[front.fx][dx1][dy1]){if(mp[dx1][dy1]=='F')return front.step+1;vis[front.fx][dx1][dy1]=true;rear.x=dx1,rear.y=dy1,rear.fx=front.fx,rear.step=front.step+1;q.push(rear);}if(ok(dx2,dy2)&&!vis[fx][dx2][dy2]){if(mp[dx2][dy2]=='F')return front.step+1;vis[fx][dx2][dy2]=true;rear.x=dx2,rear.y=dy2,rear.fx=fx,rear.step=front.step+1;q.push(rear);}}
}
void init()
{cin>>n>>m;getchar();for(int i=0; i<n; i++)gets(mp[i]);for(int i=0; i<n; i++)for(int j=0; j<m; j++)if(mp[i][j]=='S'){printf("%d\n",bfs(i,j));return ;}
}
int main()
{
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);
#endifint t;scanf("%d",&t);while(t--){init();}return 0;
}