poj1573 模拟

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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

Output

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)






没什么可说的，做不出来可以考虑退出ACM了

#include<stdio.h>
#include<string.h>
char map[100][100];
int vis[100][100];
int flag;
int x,y,h;
void dfs(int tx,int ty){
if(flag==1)
return;
if(map[tx][ty]==‘N‘)
tx-=1;
else if(map[tx][ty]==‘S‘)
tx+=1;
else if(map[tx][ty]==‘W‘)
ty-=1;
else if(map[tx][ty]==‘E‘)
ty+=1;
if(tx<1||tx>x||ty<1||ty>y)
return;
if(vis[tx][ty]==0){
vis[tx][ty]=1;
dfs(tx,ty);
}
else if(vis[tx][ty]==1){
vis[tx][ty]=2;
dfs(tx,ty);
}
else {
flag=1;
return;
}
}
int main(){

while(scanf("%d%d%d",&x,&y,&h)!=EOF){
if(x==0&&y==0&&h==0)
break;
memset(vis,0,sizeof(vis));
memset(map,0,sizeof(map));
getchar();
for(int i=1;i<=x;i++)
gets(map[i]+1);
int tx=1,ty=h;
flag=0;
vis[tx][ty]=1;
dfs(tx,ty);
int sum1=0,sum2=0;
for(int i=1;i<=x;i++){
for(int j=1;j<=y;j++){
if(vis[i][j]==1)
sum1++;
else if(vis[i][j]==2)
sum2++;
}
}
if(sum2==0)
printf("%d step(s) to exit\n",sum1);
else
printf("%d step(s) before a loop of %d step(s)\n",sum1,sum2);
}
return 0;
}

poj1573 模拟

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原文地址：http://www.cnblogs.com/13224ACMer/p/4628792.html