POJ 1573 && hdu 1035（简单模拟）

POJ 1573 && hdu 1035(简单模拟)

感谢kb神

题目链接：http://poj.org/problem?id=1573

http://acm.hdu.edu.cn/showproblem.php?pid=1035

Robot Motion

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7893		Accepted: 3813

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

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.
Output

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.
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)

#include<stdio.h>
#include<string.h>

char map[15][15];
int dis[15][15];

int main()
{
int n,m,s;
int i,j;
int si,sj;
int step;
int step1;
while(scanf("%d%d%d%*c",&n,&m,&s)!=EOF && m)
{
step=0;step1=0;
memset(dis,0,sizeof(dis));
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
scanf("%c",&map[i][j]);
getchar();
}
si=1;sj=s;
while(si>=1 && si<=n && sj>=1 && sj<=m &&!dis[si][sj])
{
step++;
dis[si][sj]=1;
if(map[si][sj]=='W')
sj-=1;
else if(map[si][sj]=='E')
sj+=1;
else if(map[si][sj]=='S')
si+=1;
else if(map[si][sj]=='N')
si-=1;
}
if(!(si>=1 && si<=n) || !(sj>=1 && sj<=m))
printf("%d step(s) to exit\n",step);
else
{
memset(dis,0,sizeof(dis));
while(!dis[si][sj])
{
step1++;
dis[si][sj]=1;
if(map[si][sj]=='W')
sj-=1;
else if(map[si][sj]=='E')
sj+=1;
else if(map[si][sj]=='S')
si+=1;
else if(map[si][sj]=='N')
si-=1;
}
step=step-step1;
printf("%d step(s) before a loop of %d step(s)\n",step,step1);
}
}
return 0;
}