Codeforces Round #354 (Div. 2) D. Theseus and labyrinth bfs
D. Theseus and labyrinth
题目连接：
http://www.codeforces.com/contest/676/problem/D
Description
Theseus has just arrived to Crete to fight Minotaur. He found a labyrinth that has a form of a rectangular field of size n × m and consists of blocks of size 1 × 1.
Each block of the labyrinth has a button that rotates all blocks 90 degrees clockwise. Each block rotates around its center and doesn't change its position in the labyrinth. Also, each block has some number of doors (possibly none). In one minute, Theseus can either push the button in order to rotate all the blocks 90 degrees clockwise or pass to the neighbouring block. Theseus can go from block A to some neighbouring block B only if block A has a door that leads to block B and block B has a door that leads to block A.
Theseus found an entrance to labyrinth and is now located in block (xT, yT) — the block in the row xT and column yT. Theseus know that the Minotaur is hiding in block (xM, yM) and wants to know the minimum number of minutes required to get there.
Theseus is a hero, not a programmer, so he asks you to help him.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and the number of columns in labyrinth, respectively.
Each of the following n lines contains m characters, describing the blocks of the labyrinth. The possible characters are:
«+» means this block has 4 doors (one door to each neighbouring block);
«-» means this block has 2 doors — to the left and to the right neighbours;
«|» means this block has 2 doors — to the top and to the bottom neighbours;
«^» means this block has 1 door — to the top neighbour;
«>» means this block has 1 door — to the right neighbour;
«<» means this block has 1 door — to the left neighbour;
«v» means this block has 1 door — to the bottom neighbour;
«L» means this block has 3 doors — to all neighbours except left one;
«R» means this block has 3 doors — to all neighbours except right one;
«U» means this block has 3 doors — to all neighbours except top one;
«D» means this block has 3 doors — to all neighbours except bottom one;
«*» means this block is a wall and has no doors.
Left, right, top and bottom are defined from representing labyrinth as a table, where rows are numbered from 1 to n from top to bottom and columns are numbered from 1 to m from left to right.
Next line contains two integers — coordinates of the block (xT, yT) (1 ≤ xT ≤ n, 1 ≤ yT ≤ m), where Theseus is initially located.
Last line contains two integers — coordinates of the block (xM, yM) (1 ≤ xM ≤ n, 1 ≤ yM ≤ m), where Minotaur hides.
It's guaranteed that both the block where Theseus starts and the block where Minotaur is hiding have at least one door. Theseus and Minotaur may be initially located at the same block.
Output
If Theseus is not able to get to Minotaur, then print -1 in the only line of the output. Otherwise, print the minimum number of minutes required to get to the block where Minotaur is hiding.
Sample Input
2 2
+*
*U
1 1
2 2
Sample Output
-1
Hint
题意
给你一个n*m的地图，然后地图有很多标志如题所述
然后他每秒钟要么可以穿过门，要么可以使得所有门都顺时针转动90°
如果你要从A到B，那么从B也必须能够到达A
问你从起点到终点的最短时间是多少
题解：
直接BFS好啦，直接暴力……
请叫我暴力大师QSC，写了700行 233
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int n,m,x1,y1,x2,y2;
char a[maxn][maxn];
int vis[maxn][maxn][4];
struct node
{
int x,y,z,time;
};
bool check(node z,node p)
{
if(p.x<1||p.x>n)return false;
if(p.y<1||p.y>m)return false;
if(a[p.x][p.y]=='*')return false;
if(a[p.x][p.y]=='-')
{
if(p.z%2==0)
{if(z.x==p.x-1||z.x==p.x+1)return false;}
else
{if(z.y==p.y-1||z.y==p.y+1)return false;}
}
if(a[p.x][p.y]=='|')
{
if(p.z%2==1)
{if(z.x==p.x-1||z.x==p.x+1)return false;}
else
{if(z.y==p.y-1||z.y==p.y+1)return false;}
}
if(a[p.x][p.y]=='^')
{
if(p.z%4==0)
if(z.x!=p.x-1)return false;
if(p.z%4==1)
if(z.y!=p.y+1)return false;
if(p.z%4==2)
if(z.x!=p.x+1)return false;
if(p.z%4==3)
if(z.y!=p.y-1)return false;
}
if(a[p.x][p.y]=='>')
{
if(p.z%4==3)
if(z.x!=p.x-1)return false;
if(p.z%4==0)
if(z.y!=p.y+1)return false;
if(p.z%4==1)
if(z.x!=p.x+1)return false;
if(p.z%4==2)
if(z.y!=p.y-1)return false;
}
if(a[p.x][p.y]=='v')
{
if(p.z%4==2)
if(z.x!=p.x-1)return false;
if(p.z%4==3)
if(z.y!=p.y+1)return false;
if(p.z%4==0)
if(z.x!=p.x+1)return false;
if(p.z%4==1)
if(z.y!=p.y-1)return false;
}
if(a[p.x][p.y]=='<')
{
if(p.z%4==1)
if(z.x!=p.x-1)return false;
if(p.z%4==2)
if(z.y!=p.y+1)return false;
if(p.z%4==3)
if(z.x!=p.x+1)return false;
if(p.z%4==0)
if(z.y!=p.y-1)return false;
}
if(a[p.x][p.y]=='L')
{
if(p.z%4==1)
if(z.x==p.x-1)return false;
if(p.z%4==2)
if(z.y==p.y+1)return false;
if(p.z%4==3)
if(z.x==p.x+1)return false;
if(p.z%4==0)
if(z.y==p.y-1)return false;
}
if(a[p.x][p.y]=='R')
{
if(p.z%4==3)
if(z.x==p.x-1)return false;
if(p.z%4==0)
if(z.y==p.y+1)return false;
if(p.z%4==1)
if(z.x==p.x+1)return false;
if(p.z%4==2)
if(z.y==p.y-1)return false;
}
if(a[p.x][p.y]=='U')
{
if(p.z%4==0)
if(z.x==p.x-1)return false;
if(p.z%4==1)
if(z.y==p.y+1)return false;
if(p.z%4==2)
if(z.x==p.x+1)return false;
if(p.z%4==3)
if(z.y==p.y-1)return false;
}
if(a[p.x][p.y]=='D')
{
if(p.z%4==2)
if(z.x==p.x-1)return false;
if(p.z%4==3)
if(z.y==p.y+1)return false;
if(p.z%4==0)
if(z.x==p.x+1)return false;
if(p.z%4==1)
if(z.y==p.y-1)return false;
}
if(vis[p.x][p.y][p.z])return false;
vis[p.x][p.y][p.z]=1;
return true;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)scanf("%s",a[i]+1);
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
queue<node> Q;
Q.push(node{x1,y1,0,0});
vis[x1][y1][0]=1;
int flag = 0;
node next;
while(!Q.empty())
{
node now = Q.front();
if(now.x==x2&&now.y==y2)
{
printf("%d\n",now.time);
return 0;
}
Q.pop();
if(a[now.x][now.y]=='+')
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
if(a[now.x][now.y]=='-')
{
if(now.z%2==0)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
}
else
{
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
}
if(a[now.x][now.y]=='|')
{
if(now.z%2==1)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
}
else
{
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
}
if(a[now.x][now.y]=='^')
{
if(now.z%4==0){
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==1)
{
next=now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==2)
{
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==3)
{
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
}
}
if(a[now.x][now.y]=='v')
{
if(now.z%4==2){
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==3)
{
next=now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==0)
{
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==1)
{
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
}
}
if(a[now.x][now.y]=='<')
{
if(now.z%4==1){
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==2)
{
next=now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==3)
{
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==0)
{
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
}
}
if(a[now.x][now.y]=='>')
{
if(now.z%4==3){
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==0)
{
next=now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==1)
{
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==2)
{
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
}
}
if(a[now.x][now.y]=='L')
{
if(now.z%4==0)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==1)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==2)
{
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==3)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
}
}
if(a[now.x][now.y]=='R')
{
if(now.z%4==2)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==3)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==0)
{
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==1)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
}
}
if(a[now.x][now.y]=='U')
{
if(now.z%4==3)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==0)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==1)
{
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==2)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
}
}
if(a[now.x][now.y]=='D')
{
if(now.z%4==1)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==2)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==3)
{
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x++;
next.time++;
if(check(now,next))
Q.push(next);
}
else if(now.z%4==0)
{
next = now;
next.y++;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.y--;
next.time++;
if(check(now,next))
Q.push(next);
next=now;
next.x--;
next.time++;
if(check(now,next))
Q.push(next);
}
}
next=now;
next.time++;
next.z=(next.z+1)%4;
if(!vis[next.x][next.y][next.z])
{
vis[next.x][next.y][next.z]=1;
Q.push(next);
}
}
printf("-1\n");
}
/*
3 3
->v
*+|
+*^
3 3
1 1
*/Codeforces Round #354 (Div. 2) D. Theseus and labyrinth bfs的更多相关文章
- Codeforces Round #354 (Div. 2) D. Theseus and labyrinth
题目链接: http://codeforces.com/contest/676/problem/D 题意: 如果两个相邻的格子都有对应朝向的门,则可以从一个格子到另一个格子,给你初始坐标xt,yt,终 ...
- Codeforces Round #354 (Div. 2) ABCD
Codeforces Round #354 (Div. 2) Problems # Name A Nicholas and Permutation standard input/out ...
- Codeforces Round #354 (Div. 2)-D
D. Theseus and labyrinth 题目链接:http://codeforces.com/contest/676/problem/D Theseus has just arrived t ...
- Codeforces Round #354 (Div. 2)
贪心 A Nicholas and Permutation #include <bits/stdc++.h> typedef long long ll; const int N = 1e5 ...
- Codeforces Round #354 (Div. 2)-C
C. Vasya and String 题目链接:http://codeforces.com/contest/676/problem/C High school student Vasya got a ...
- Codeforces Round #354 (Div. 2)-B
B. Pyramid of Glasses 题目链接:http://codeforces.com/contest/676/problem/B Mary has just graduated from ...
- Codeforces Round #354 (Div. 2)-A
A. Nicholas and Permutation 题目链接:http://codeforces.com/contest/676/problem/A Nicholas has an array a ...
- Codeforces Round #354 (Div. 2) C. Vasya and String
题目链接: http://codeforces.com/contest/676/problem/C 题解: 把连续的一段压缩成一个数,对新的数组求前缀和,用两个指针从左到右线性扫一遍. 一段值改变一部 ...
- Codeforces Round #354 (Div. 2)_Vasya and String(尺取法)
题目连接:http://codeforces.com/contest/676/problem/C 题意:一串字符串,最多改变k次,求最大的相同子串 题解:很明显直接尺取法 #include<cs ...
随机推荐
- mysql备份的 三种方式【转】
备份的本质就是将数据集另存一个副本,但是原数据会不停的发生变化,所以利用备份只能回复到数据变化之前的数据.那变化之后的呢?所以制定一个好的备份策略很重要. 一.备份的目的 做灾难恢复:对损坏的数据进行 ...
- STM32 IAP升级
STM32 IAP在线升级,用Jlink设置读保护后前5K字节是默认加了写保护的,导致IAP升级时擦除和写入FLASH不成功,可以做两个boot,前5k为第一个boot程序,上电时负责跳转到APP还是 ...
- Failed to load class "org.slf4j.impl.StaticLoggerBinder"
调试程序出现如下错误: SLF4J: Failed to load class "org.slf4j.impl.StaticLoggerBinder".SLF4J: Default ...
- XP远程连接Win10，提示【远程计算机需要网络级别身份验证，而您的计算机不支持该验证】
最近电脑安装了Win10系统,在办公室可以通过其他电脑远程,但是回去后使用自己的电脑(XP系统)进行远程提示失败, 提示[远程计算机需要网络级别身份验证,而您的计算机不支持该验证],然后上网查找资料, ...
- IOS使用xcode编译代码
一.安装xcode 在app store中搜索xcode然后点击安装即可. 二.创建第一个app 1.启动xcode,单击Lauchpad 2.单击xcode启动 3.单击“Create a new ...
- elasticsearch文档学习
1.集群 节点(一个elasticsearch实体) 索引 主节点 :集群级别变更,新增或移除节点,索引: 主节点不参与文档级别搜索和变更. 分片(shard):一个完整的搜索引擎,lucene ...
- WinScp几个极大提高开发效率的小功能
WinSCP 是一个 Windows 环境下使用 SSH 的开源图形化 SFTP 客户端.同时支持 SCP 协议.它的主要功能就是在本地与远程计算机间安全的复制文件. 最近研究了一下winscp的一些 ...
- Sqlserver双机热备文档(无域)
1. 配制环境 OS:Win7 DB:SQL Server R2 2. 基本配制 1. 开启sqlServer服务如下图-1 图-1 2. 开启sqlServer的tcp/i ...
- sicily 1198. Substring （递归全排列+排序）
DescriptionDr lee cuts a string S into N pieces,s[1],…,s[N]. Now, Dr lee gives you these N sub-strin ...
- OpenCV处理直方图
直方图可以用来描述各种不同的事物,如物体的色彩分布.物体边缘梯度模板,以及表示目标位置的当前假设. 简单的说,直方图就是对数据进行统计,将统计值组织到一系列事先定义好的bin中.bin中的数值是从数据 ...