Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3266 Accepted Submission(s): 1519
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.
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.
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.
3 6 5 NEESWE WWWESS SNWWWW 4 5 1 SESWE EESNW NWEEN EWSEN 0 0
10 step(s) to exit 3 step(s) before a loop of 8 step(s)
#include<cstdio> #include<algorithm> #include<cstring> #define N 200 using namespace std; char map[N][N]; int m,n,begin; int step[N][N]; bool s[N][N]; int _step,loop; void dfs(int x,int y) { if(map[x][y]=='W') { if(y-1<0) { loop=0;_step=step[x][y]+1; } else if(s[x][y-1]) { _step=step[x][y-1]; loop=step[x][y]-step[x][y-1]+1; return; } else { s[x][y-1]=1; step[x][y-1]=step[x][y]+1; dfs(x,y-1); } } if(map[x][y]=='E') { if(y+1>=n) { loop=0;_step=step[x][y]+1; return; } else if(s[x][y+1]) { loop=step[x][y]-step[x][y+1]+1; _step=step[x][y+1]; return; } else { s[x][y+1]=1; step[x][y+1]=step[x][y]+1; dfs(x,y+1); } } if(map[x][y]=='N') { if(x-1<0) { loop=0;_step=step[x][y]+1; return; } else if(s[x-1][y]) { loop=step[x][y]-step[x-1][y]+1; _step=step[x-1][y]; return; } else { s[x-1][y]=1; step[x-1][y]=step[x][y]+1; dfs(x-1,y); } } if(map[x][y]=='S') { if(x+1>=m) { loop=0;_step=step[x][y]+1; return; } else if(s[x+1][y]) { _step=step[x+1][y]; loop=step[x][y]-step[x+1][y]+1; return; } else { s[x+1][y]=1; step[x+1][y]=step[x][y]+1; dfs(x+1,y); } } } int main() { //freopen("4.txt","r",stdin); while(scanf("%d%d%d",&m,&n,&begin)&&(m||n||begin)) { memset(map,-1,sizeof(map)); memset(step,0,sizeof(step)); memset(s,0,sizeof(s)); for(int i=0;i<m;i++) scanf("%s",map[i]); s[0][begin-1]=1; dfs(0,begin-1); if(loop) printf("%d step(s) before a loop of %d step(s)\n",_step,loop); else printf("%d step(s) to exit\n",_step); } return 0; }