hdoj 1035 Robot Motion
2015-07-01 22:41
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Robot Motion
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7974 Accepted Submission(s):
3685
[align=left]Problem Description[/align]
![](http://acm.hdu.edu.cn/data/images/1035-1.gif)
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
3 6 5
NEESWE
WWWESS
SNWWWW
4 5 1
SESWE
EESNW
NWEEN
EWSEN
0 0
[align=left]Sample Output[/align]
10 step(s) to exit
3 step(s) before a loop of 8 step(s)
做题时两个地方被坑到:
1、刚开始的时候横纵坐标搞反了
2、输出的结果错误后来发现输入的时候数组是从0开始的而在搜索判断中却是从数组中1 开始判断
#include<stdio.h> #include<string.h> #define MAX 110 char map[MAX][MAX];//所要走的地图 int vis[MAX][MAX];//用来记录当前走了多少步 int n,m,t; int x2,y2;//用于记录当前步的上一步 void dfs(int x1,int y1) { if(x1>n||x1<1||y1<1||y1>m)//如果超出地图范围则证明已经走出去了 { printf("%d step(s) to exit\n",vis[x2][y2]); return ; } else if(vis[x1][y1])//如果不为0则证明此处已经走过形成环 { printf("%d step(s) before a loop of %d step(s)\n",vis[x1][y1]-1,vis[x2][y2]-vis[x1][y1]+1); return ; } vis[x1][y1]=vis[x2][y2]+1;//当前走的步数是上一步加1 x2=x1;y2=y1; if(map[x1][y1]=='W')//向左走 y1-=1; else if(map[x1][y1]=='S')//向下走 x1+=1; else if(map[x1][y1]=='E')//向右走 y1+=1; else if(map[x1][y1]=='N')//向上走 x1-=1; dfs(x1,y1); } int main() { int j,i,s,k; int x1,x2,y1,y2; while(scanf("%d%d",&n,&m)&&n!=0&&m!=0) { scanf("%d",&t); for(i=1;i<=n;i++) { getchar(); for(j=1;j<=m;j++) { scanf("%c",&map[i][j]); } } x1=1;y1=t;//起点 x2=x1;y2=y1; memset(vis,0,sizeof(vis));//数组清零 dfs(x1,y1); } return 0; }
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