uva 10452 Marcus
来源:互联网 发布:云计算数据中心建设 编辑:程序博客网 时间:2024/06/05 00:17
Problem I
Marcus, help!
Input: standard input
Output: standard output
Time Limit: 2 Seconds
"First, the breath of God.
Only the penitent man will pass.
Second, the Word of God,
Only in the footsteps of God will he proceed.
Third, the Path of God,
Only in the leap from the lion's head will he prove his worth."
(taken from the movie "Indiana Jones and the Last Crusade", 1989)
To get to the grail, Indiana Jones needs to pass three challenges. He successfully masters the first one and steps up to the second. A cobblestone path lies before him, each cobble is engraved with a letter. This is the second challenge, the Word of God, the Name of God. Only the cobbles having one of the letters "IEHOVA" engraved can be stepped on safely, the ones having a different letter will break apart and leave a hole.
Unfortunately, he stumbles and falls into the dust, and the dust comes into his eyes, making it impossible for him to see. So he calls for Marcus Brody and asks Marcus to tell him in which
direction to go to safely reach the other side of the cobblestone path. Because of the dust in his eyes, Indy only can step "forth" to the stone right in front of him or do a side-step to the stone on the "left" or the "right" side of him. These ("forth", "left", "right") are also the commands Marcus gives to him.
Input
The first line of the input contains a single number indicating the number of test cases that follow.
Each test case starts with a line containing two numbers m and n (2 <= m, n <= 8), the length m and the width n of the cobblestone path. Then follow m lines, each containing n characters ('A' to 'Z', '@', '#'), the engravement of the respective cobblestone. Indy's starting position is marked with the '@' character in the last line, the destination with the character '#' in the first line of the cobblestone path.
Each of the letters in "IEHOVA" and the characters '@' and '#' appear exactly once in each test case. There will always be exactly one path from Indy's starting position via the stones with the letters "IEHOVA" engraved on (in that order) to the destination. There will be no other way to safely reach the destination.
Output
For each test case, output a line with the commands Marcus gives to Indy so that Indy safely reaches the other side. Separate two commands by one space character.
Sample Input
2
6 5
PST#T
BTJAS
TYCVM
YEHOF
XIBKU
N@RJB
5 4
JA#X
JVBN
XOHD
DQEM
T@IY
Sample Output
forth forth right right forth forth forth
right forth forth left forth forth right
#include <iostream>#include <cstdio>using namespace std;const int d[3][2]={{0,-1},{-1,0},{0,1}};const char s[8]={'@','I','E','H','O','V','A','#'};const char *t[]={"left","forth","right"};const int maxn=10;char a[maxn][maxn];bool visited[maxn][maxn];int n,m,p,q;void input(){ scanf("%d %d",&n,&m); for(int i=0;i<n;i++) { getchar(); for(int j=0;j<m;j++) { scanf("%c",&a[i][j]); if(a[i][j]=='@') p=i,q=j; } }}bool judge(int x,int y){ if(x>=0 && x<n && y>=0 && y<m) return true; return false;}void dfs(int x,int y,int depth){ for(int i=0;i<3;i++) { int xx=x+d[i][0],yy=y+d[i][1]; if(judge(xx,yy) && a[xx][yy]==s[depth]) { printf("%s",t[i]); if(depth<=6) printf(" "); dfs(xx,yy,depth+1); } }}int main(){ int T; scanf("%d",&T); while(T--) { input(); dfs(p,q,1); printf("\n"); } return 0;}
- UVa, 10452 Marcus
- uva 10452 Marcus
- UVa 10452 - Marcus
- UVa 10452 - Marcus
- UVA 10452 Marcus
- UVA 10452 Marcus, help!( DFS )
- UVa10452 - Marcus
- Marcus McNeill's leverage
- dfs-Marcus, help!-uva10452
- Marcus在召集PHP开发工具的志愿者
- uva 10452
- O'Reilly数据秀播客|Adam Marcus谈智能系统和人工辅助计算
- 视频出炉 | LeCun、Marcus激辩AI是否需要类似人类的认知能力
- Yann LeCun vs Gary Marcus:AI 学习的未来更多取决于先天还是后天?
- uva
- UVA
- UVA
- UVA
- android 学习_Fragment 介绍
- 统计但是还没有进行重定向和fopen
- submit提交后怎么刷新页面
- memcached 分布式取决于客户端
- vTiger CRM 6.1.0 索引篇
- uva 10452 Marcus
- 第十七章 17.5.2节练习 & 17.5.3节练习
- vector,CCArray等元删除满足条件元素的删除(比如删除大于2的元素)
- lotusphp里面和thinkphp里面的C方法重复怎么解决
- Android学习笔记 - 目录
- uvaoj 1364 - Knights of the Round Table
- cookie 赋了值但是不起作用 需要先清空再赋值就可以
- 水晶报表打印乱码问题
- 设置ExpandableListView右边的箭头