hdu-1573 Robot Motion
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 10219 | Accepted: 4977 |
Description

A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a grid. The possible instructions are
N north (up the page)
S south (down the page)
E east (to the right on the page)
W west (to the left on the page)
For example, suppose the robot starts on the north (top) side of Grid 1 and starts south (down). The path the robot follows is shown. The robot goes through 10 instructions in the grid before leaving the grid.
Compare what happens in Grid 2: the robot goes through 3 instructions only once, and then starts a loop through 8 instructions, and never exits.
You are to write a program that determines how long it takes a robot to get out of the grid or how the robot loops around.
Input
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.
Output
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.
Sample Input
3 6 5
NEESWE
WWWESS
SNWWWW
4 5 1
SESWE
EESNW
NWEEN
EWSEN
0 0 0
Sample Output
10 step(s) to exit
3 step(s) before a loop of 8 step(s)
Source
//主要是推断是否构成循环。依据题目特点。能够设置一个标记数组,假设走过了就不能在走,直到退出循环。
用另外一个数组表示到这一个坐标的步数。那么退出循环的时候事实上已经到曾经走过的点。输出步数,然后循环节就是总步数减去当前步数。
#include<stdio.h>
#include<string.h>
int main()
{
int n,m,k,l,p,step;
char s[11][11];
int map[11][11];
while(scanf("%d %d",&n,&m)!=EOF)
{
if(n==0||m==0)break;
scanf("%d",&k);
for(int i=0;i<n;i++)
scanf("%s",s[i]);
l=0;
p=k-1;
step=0;
while(l>=0&&l<n&&p>=0&&p<m&&s[l][p]!='0')
{
if(s[l][p]=='N')
{
s[l][p]='0';
map[l][p]=++step; //先计算当前步数,然后在改变坐标值
l--;
}
else if(s[l][p]=='S')
{
s[l][p]='0';
map[l][p]=++step;
l++;
}
else if(s[l][p]=='E')
{
s[l][p]='0';
map[l][p]=++step;
p++;
}
else if(s[l][p]=='W')
{
s[l][p]='0';
map[l][p]=++step;
p--;
}
}
if(s[l][p]=='0') //表示循环 了、
printf("%d step(s) before a loop of %d step(s)\n",map[l][p]-1,step+1-map[l][p]);
else printf("%d step(s) to exit\n",step);
}
return 0;
}
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