Jury Jeopardy （这是一道单纯的模拟题）
What would a programming contest be without a problem featuring an ASCII-maze? Do not despair: one of the judges has designed such a problem.
The problem is about a maze that has exactly one entrance/exit, contains no cycles and has no empty space that is completely enclosed by walls. A robot is sent in to explore the entire maze. The robot always faces the direction it travels in. At every step, the robot will try to turn right. If there is a wall there, it will attempt to go forward instead. If that is not possible, it will try to turn left. If all three directions are unavailable, it will turn back.
The challenge for the contestants is to write a program that describes the path of the robot, starting from the entrance/exit square until it finally comes back to it. The movements are described by a single letter: 'F' means forward, 'L' is left, 'R' is right and 'B' stands for backward. Each of 'L', 'R' and 'B' does not only describe the change in orientation of the robot, but also the advancement of one square in that direction. The robot’s initial direction is East. In addition, the path of the robot always ends at the entrance/exit square.
The judge responsible for the problem had completed all the samples and testdata, when disaster struck: the input file got deleted and there is no way to recover it! Fortunately the output and the samples are still there. Can you reconstruct the input from the output? For your convenience, he has manually added the number of test cases to both the sample output and the testdata output.
On the first line one positive number: the number of test cases. After that per test case:
one line with a single string: the movements of the robot through the maze.
On the first line one positive number: the number of test cases, at most 100. After that per test case:
one line with two space-separated integers h and w (3 ≤ h, w ≤ 100): the height and width of the maze, respectively.
h lines, each with w characters, describing the maze: a '#' indicates a wall and a '.' represents an empty square.
The entire contour of the maze consists of walls, with the exception of one square on the left: this is the entrance. The maze contains no cycles (i.e. paths that would lead the robot back to a square it had left in another direction) and no empty squares that cannot be reached from the entrance. Every row or column --- with the exception of the top row, bottom row and right column --- contains at least one empty square.
3
FFRBLF
FFRFRBRFBFRBRFLF
FRLFFFLBRFFFRFFFRFRFBRFLBRFRLFLFFR
3
4 4
####
...#
##.#
####
7 5
#####
...##
##.##
#...#
##.##
##.##
#####
7 7
#######
#...#.#
#.#...#
#.#.###
..###.#
#.....#
#######
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
#define PY 100
#define INF 0x3f3f3f3f
bool book[][];
char ch[];
int main()
{
int n; scanf("%d",&n); printf("%d\n",n);
while(n--)
{
memset(book,,sizeof(book));
getchar();
scanf("%s",ch);
int i=;
int nowx=+PY;
int nowy=+PY;
int miny=PY,maxy=PY,maxx=PY,minx=PY;
int t=;//东
book[nowx][nowy]=; while(ch[i]!='\0')
{
if(ch[i]=='F')
{
if(t==)
{
nowy++; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
if(t==)
{
nowx++; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
if(t==)
{
nowy--; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
if(t==)
{
nowx--; book[nowx][nowy]=;
// printf("%d %d\n",nowx,nowy);
} }
if(ch[i]=='B')
{
if(t==)
{
t=;
nowy--; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
else if(t==)
{
t=;
nowx--; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy); }
else if(t==)
{
t=;
nowy++; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
else if(t==)
{
t=;
nowx++; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
}
if(ch[i]=='R')
{
if(t==)
{
t=;//南
nowx++; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
else if(t==)
{
t=;//西
nowy--; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
else if(t==)
{
t=;//北
nowx--; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
else if(t==)
{
t=;
nowy++; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
}
if(ch[i]=='L')
{
if(t==)
{
t=;
nowx--; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
else if(t==)
{
t=;
nowy++; book[nowx][nowy]=;
// printf("%d %d\n",nowx,nowy);
}
else if(t==)
{
t=;
nowx++; book[nowx][nowy]=;
// printf("%d %d\n",nowx,nowy);
}
else if(t==)
{
t=;
nowy--; book[nowx][nowy]=;
//printf("%d %d\n",nowx,nowy);
}
}
i++;
minx=min(minx,nowx);
maxx=max(maxx,nowx);
miny=min(miny,nowy);
maxy=max(maxy,nowy);
}
int m=maxy-miny++;int n=maxx-minx++; printf("%d %d\n",n,m);
for(int i=minx- ; i<=maxx+ ;i++)
{
for(int j=miny ;j<=maxy+ ; j++)
if(book[i][j]==)
putchar('.');
else
putchar('#');
puts("");
} }
return ;
}
这是大佬的代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int net[][]={{,},{,},{,-},{-,}};
bool book[][];
char ch[];
int main()
{
int t;
scanf("%d",&t);
printf("%d\n",t);
while(t--)
{
memset(book,,sizeof(book));
scanf("%s",ch);
int n=strlen(ch);
int r=;
int nowx,nowy,maxy,miny,maxx,minx;
nowx=nowy=maxy=miny=maxx=minx=;
book[nowx][nowy]=;
for(int i= ; i<n ; i++)
{
if(ch[i]=='F')
{
nowx+=net[r][];
nowy+=net[r][];
book[nowx][nowy]=; }
else if(ch[i]=='R')
{
r=(r+)%;
nowx+=net[r][];
nowy+=net[r][];
book[nowx][nowy]=;
}
else if(ch[i]=='B')
{
r=(r+)%;
nowx+=net[r][];
nowy+=net[r][];
book[nowx][nowy]=;
}
else if(ch[i]=='L')
{
r=(r+)%;
nowx+=net[r][];
nowy+=net[r][];
book[nowx][nowy]=;
}
maxx=max(nowx,maxx);
minx=min(nowx,minx);
maxy=max(nowy,maxy);
miny=min(nowy,miny);
}
printf("%d %d\n",maxx-minx+,maxy-miny+);
for(int i=minx- ; i<=maxx+ ; i++)
{
for(int j=miny ; j<=maxy+ ; j++)
if(book[i][j]==)
printf(".");
else
printf("#");
printf("\n");
} } return ;
}
看到了吗，同样的一道问题的区别，虽然我的代码快一点点，然并软。
来简单分析下大神与我的区别1：我是根据方向来确定下一点。大神是根据点来确定方向，这就是个本质的区别，还是要多学习人家好的代码
最新文章
- Servlet监听器笔记总结
- quartz CronExpression表达式
- 19.Java 注解
- 12款非常精致的免费 HTML5 & CSS3 网站模板
- 搜索引擎LuceneNet
- safari浏览器添加书签的方法
- [安卓] 9、线程、VIEW、消息实现从TCP服务器获取数据动态加载显示
- Buffer cache 的调整与优化
- Fedora16 安装相关
- 【Android平台安全方案】の ＃00-请不要在外部存储（SD卡）加密存储的敏感信息
- Oracle数据库部分迁至闪存存储方案
- ORACLE窗口函数
- Spring源码追踪1——doGetBean（为什么org.springframework.data.redis.core.RedisTemplate的实例可以注入为ListOperations）
- Python中For循环
- python 爆破
- SNF开发平台WinForm-Grid表格控件大全
- 强大的JQuery表单验证插件 FormValidator使用介绍
- es分布式文档系统_bulk api的奇特json格式与底层性能优化关系
- Java基础 【Arrays 类的使用】
- Gravitational Teleport docker-compose简单运行