| Robot Motion | ||||||
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| Description | ||||||
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| Input | ||||||
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There will be one or more grids for robots to navigate. The data for each is in the following form. On the first line are three integers separated by blanks: the number of rows in the grid, the number of columns in the grid, and
the number of the column in which the robot enters from the north. The possible entry columns are numbered starting with one at the left. Then come the rows of the direction instructions. Each grid will have at least one and at most 10 rows and columns of
instructions. The lines of instructions contain only the characters N, S, E, or W with no blanks. The end of input is indicated by a row containing 0 0 0.
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| Output | ||||||
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For each grid in the input there is one line of output. Either the robot follows a certain number of instructions and exits the grid on any one the four sides or else the robot follows the instructions on a certain number of locations
once, and then the instructions on some number of locations repeatedly. The sample input below corresponds to the two grids above and illustrates the two forms of output. The word "step" is always immediately followed by "(s)" whether or not the number before
it is 1.
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| Sample Input | ||||||
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3 6 5 NEESWE WWWESS SNWWWW 4 5 1 SESWE EESNW NWEEN EWSEN 0 0 0 | ||||||
| Sample Output | ||||||
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10 step(s) to exit 3 step(s) before a loop of 8 step(s) |
题目大意:每一个gird(格子)上都标记有下一步要走到哪里,问如果能够走出n*m大的地图,就按照样例1输出,否则如果走出了环,先输出在走到环起点之前走了多少步,然后输出在环里能重复走的步数,多组输入,0 0 0break;
思路:简单模拟,牢记初始化,注意起点的标记。
AC代码:
#include<stdio.h>
#include<string.h>
using namespace std;
int output[1050][1050];
char a[1050][1050];
//n-up 1 13
//s-down 2 18
//e-right 3 4
//w-left 4 22
int n,m,t;
void solve()
{
memset(output,0,sizeof(output));
output[0][t]=1;
int flag=0;
int ans=0;
int step=1;
int x=0;
int y=t;
while(1)
{
//printf("%d %d %d\n",x,y,output[x][y]);
int cur=a[x][y]-'A';
if(cur==13)
{
x-=1;
}
if(cur==18)
{
x+=1;
}
if(cur==4)
{
y+=1;
}
if(cur==22)
{
y-=1;
}
if(x<0||y<0||x>=n||y>=m)
break;
step++;
if(output[x][y]==0)
output[x][y]=step;
else
{
//printf("%d %d\n",x,y);
ans=output[x][y];
flag=step-output[x][y];
break;
}
}
if(flag==0)
printf("%d step(s) to exit\n",step);
else
printf("%d step(s) before a loop of %d step(s)\n",ans-1,flag);
}
int main()
{
while(~scanf("%d%d%d",&n,&m,&t))
{
if(n==0&&m==0&&t==0)break;
for(int i=0;i<n;i++)
{
scanf("%s",a[i]);
}
t--;
solve();
}
}
本文探讨了机器人在网格中导航的问题,通过一系列指令决定机器人是否能顺利离开网格,或是进入循环并重复执行某些指令。文章详细介绍了算法实现步骤及输入输出格式,包括如何模拟机器人的移动过程以及判断其行为的两种可能结果。



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