Description
![](http://poj.org/images/1573_1.jpg)
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.
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 Output
10 step(s) to exit
3 step(s) before a loop of 8 step(s)
题意: 不难理解, 我简单说说, 给出r行c列, 然后从第一行的第i个位置开始走
每个位置有标记E W S N, 分别表示向东, 向西, 向南, 向北
然后, 有两种情况, 一种能走出去, 一种是陷入死循环
做法: 每一步都标上是第几步走的就OK了~ 然后按题意模拟~
AC代码:
#include<stdio.h>
char str[100][100];
int main() {
int r, c, in;
while(scanf("%d %d %d", &r, &c, &in) != EOF) {
if(r == 0 && c ==0 && in == 0)
break;
getchar();
for(int i = 0; i < r; i++)
gets(str[i]);
int tr, tc, loop;
int step = 0;
int mark = 1;
tr = 0; tc = in-1;
while(1) {
if(str[tr][tc] == 'E') {
step++;
str[tr][tc] = step + '0';
if(tc == c-1)
break;
else
tc = tc + 1;
}
else if(str[tr][tc] == 'W') {
step++;
str[tr][tc] = step + '0';
if(tc == 0)
break;
else
tc = tc - 1;
}
else if(str[tr][tc] == 'S') {
step++;
str[tr][tc] = step + '0';
if(tr == r-1)
break;
else
tr = tr + 1;
}
else if(str[tr][tc] == 'N') {
step++;
str[tr][tc] = step + '0';
if(tr == 0)
break;
else
tr = tr - 1;
}
else {
mark = 0;
loop = str[tr][tc] - '0' - 1;
printf("%d step(s) before a loop of %d step(s)
", loop, step - loop);
break;
}
}
if(mark == 1)
printf("%d step(s) to exit
", step);
}
return 0;
}