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
- asp.net 伪静态实现（UrlRewritingNet）
UrlRewritingNet.UrlRewriter源码地址 https://github.com/aspnetde/UrlRewritingNet部署步骤: 步骤一: <!--只允许存在一个 ...
- sublime Text快捷键(超级全）
sublime Text快捷键(超级全) Ctrl+Shift+P:打开命令面板 Ctrl+P:搜索项目中的文件 Ctrl+G:跳转到第几行 Ctrl+W:关闭当前打开文件 Ctrl+Shift+W: ...
- 消息 8101，级别 16，状态 1，第 1 行仅当使用了列列表并且 IDENTITY_INSERT 为 ON 时，才能为表'CUSTOMER_TBL'中的标识列指定显式值。
像这样的问题怎么解决呢? 问题分析: 意思是你的主键是自动编号类型的,所以不能向该列插入数据. 解决办法: 执行 语句 :SET IDENTITY_INSERT CUSTOMER_TBL ON 然后在 ...
- MySQL学习笔记：set autocommit=0和start transaction
在MySQL中,接触事务的例子比较少,今晚在一个简单的存储过程中,循环插入50000条id数据的时候,得知必须开事务,如果逐提交数据,将会导致速度异常的慢. SET autocommit = 0; ...
- 【LOJ】#2446. 「NOI2011」 NOI 嘉年华
题解 一道神奇的dp 我们发现关于两个东西的记录很难办,但是我们发现在固定时间区间内,如果A场地举办的活动数是一定的,那么B场地肯定举办的活动越多越好 我们预处理一个\(num[i][j]\)表示时间 ...
- Bechmarksql
使用方法 1.配置JDK环境 JDK最好是1.5以上的版本 JAVA_HOME=/usr/java/jdk1.6.0_22 PATH=$PATH:$JAVA_HOME/bin 2.准备工作 把要测试的 ...
- spring boot配置使用fastjson
一.前言 spring boot默认使用jackson来操作json数据,相比于jackson,fastjson更好用,功能也强大,所以这里记录一下在spring boot中配置使用fastjson的 ...
- php利用root权限执行shell脚本 （转）
转一篇博客,之前搞这个东西搞了好久,结果今天晚上看到了一篇救命博客,瞬间开心了...转载转载 利用sudo来赋予Apache的用户root的执行权限,下面记录一下: 利用PHP利用root权限执行sh ...
- h5行情k线开发
前言 由于公司项目需要,要做港股行情的H5版本,经过分析需求,大致有两块难点: 一是行情的推送接收,二是行情K线的生成及相关操作.本文章主要分析行情K线的相关实现,由于我们前端团队之前 ...
- Java 集合补充
集合大致可以分List,Set,Queue,Map四种体系. 集合和数组不一样,数组元素可以是基本类型的值,也可以是对象(的引用变量),集合里只能保存对象(的引用变量). 访问:如果访问List集合中 ...