1010 Robot Motion
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.
题意:题目给出一个矩阵,让机器人按照规定行走;如果最终走入一个循环中,则输出进入循环前的的步数和循环的步数,如果最终走出矩阵范围也是输行走的步数;
规则如下:
E:向右走一步;W:向左走一步;N:向上走一步;S向下走一步;
AC代码:
#include<iostream>
#include<cstring>
#include<cstdio> using namespace std; int dp[][]={};
char ch[][];
int x,y,s;
int em()
{
int number=;
int a=,b=s;
while(){
if(dp[a][b]){//循环情况的处理;
cout<<dp[a][b]-<<" step(s) before a loop of "<<number-dp[a][b]<<" step(s)"<<endl;
return ;
}
if(a==||a==x+||b==||b==y+){cout<<number-<<" step(s) to exit"<<endl;return ;}
dp[a][b]=number;
switch(ch[a][b])
{
case 'E':b++;break;
case 'W':b--;break;
case 'S':a++;break;
case 'N':a--;break;
}
number++;
}
} int main()
{
freopen("1.txt","r",stdin);
int i,j;
while(){
memset(dp,,sizeof(dp));
cin>>x>>y>>s;
if(x==y&&y==s&&s==)return ;
for(i=;i<=x;i++)
for(j=;j<=y;j++)cin>>ch[i][j];
em();
}
}
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