Problem 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
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.
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 Input
3 6 5
NEESWE
WWWESS
SNWWWW
4 5 1
SESWE
EESNW
NWEEN
EWSEN
0 0
Sample Output
10 step(s) to exit
3 step(s) before a loop of 8 step(s)
解题思路：直接用深搜。此题有两种情况：
1.一种就是能走出来；另一种就是走不出来;
2.在里面有一个循环的路径。
对于情况1很好解决，只需用路径标记数组判断没有走到走过的路并且走出啦矩形方框。而情况2也差不多，当走到走过的路时，记录此时走的步数step1，并把再从此点走一次，再次回到此点时，记录步数为step2，那么循环的步数就为step2-step1，循环前的步数就为2step1-step2。如图（红线走两次，绿线走一次）：
代码为:
#include#include #include #include using namespace std;int n,m,k,step,step1,flag;char s[100][100];int map[100][100];int f(int x,int y) //判断是否出去啦，减枝{ if(x<0||x>=n||y<0||y>=m) { return 0; } return 1;}void f1(int x,int y) //遇到啦重复点{ void dfs(int a,int b); if(map[x][y]) { if(flag==1) { step++; step1=step; flag=2; memset(map,0,sizeof(map)); map[x][y]=1; dfs(x,y); } else if(flag==2) { step++; printf("%d step(s) before a loop of %d step(s)\n",2*step1-step,step-step1); } }}void dfs(int x,int y){ if(!f(x,y)) { printf("%d step(s) to exit\n",step); } else { if(s[x][y]=='N') { if(!map[x-1][y]) { map[x-1][y]=1; step++; dfs(x-1,y); } else { f1(x-1,y); } } else if(s[x][y]=='S') { if(!map[x+1][y]) { map[x+1][y]=1; step++; dfs(x+1,y); } else { f1(x+1,y); } } else if(s[x][y]=='W') { if(!map[x][y-1]) { map[x][y-1]=1; step++; dfs(x,y-1); } else { f1(x,y-1); } } else if(s[x][y]=='E') { if(!map[x][y+1]) { map[x][y+1]=1; step++; dfs(x,y+1); } else { f1(x,y+1); } } }}int main(){ while(scanf("%d%d",&n,&m)!=EOF&&n&&m) { scanf("%d",&k); memset(map,0,sizeof(map)); for(int i=0;i