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.
InputThere 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.
OutputFor 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)
题意很简单： 机器人按照指令走方格，，能走出去输出第一种答案 不能走出去（一定是循环）输出循环前的步数和循环中的步数
思路： 方法1： 自己的理解是dfs爆搜 难点是处理循环，判断循环
方法2： 别人的理解链式前向性 因为代码我也理解了所以也写进博客了
方法一代码（dfs）：
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<set> #include<algorithm> #include<map> #define maxn 200005 using namespace std; int step=0; char mp[1000][1000]; int op=0; int n,m,k; int vis[1000][1000]={0}; int dfs(int i,int j) { if(i<0||i>=n||j<0||j>=m) { return step; } if(vis[i][j]==2){ return step-op; } if(vis[i][j]==1) { op++; if(mp[i][j]=='W'){vis[i][j]=2;dfs(i,j-1);} else if(mp[i][j]=='E'){vis[i][j]=2;dfs(i,j+1);} else if(mp[i][j]=='S'){vis[i][j]=2;dfs(i+1,j);} else if(mp[i][j]=='N'){vis[i][j]=2;dfs(i-1,j);} } if(vis[i][j]==0){ step++; if(mp[i][j]=='W'){vis[i][j]=1;dfs(i,j-1);} else if(mp[i][j]=='E'){vis[i][j]=1;dfs(i,j+1);} else if(mp[i][j]=='S'){vis[i][j]=1;dfs(i+1,j);} else if(mp[i][j]=='N'){vis[i][j]=1;dfs(i-1,j);} } } int main() { while(cin>>n>>m) { if(n==0&&m==0)break; cin>>k; getchar(); for(int i=0;i<n;i++) for(int j=0;j<m;j++) { cin>>mp[i][j]; } memset(vis,0,sizeof(vis)); step=0;op=0; int ans=dfs(0,k-1); if(op==0) { printf("%d step(s) to exit ",step); } else printf("%d step(s) before a loop of %d step(s) ",step-op,op); } return 0; }
方法2（链式前向性）：
#include<stdio.h> #include<string.h> int main() { char str[200][200]; int map[200][200]; int c,r,start,flag; int i,j; while(scanf("%d%d%d",&r,&c,&start),c!=0||r!=0||start!=0) { memset(str,0,sizeof(str)); memset(map,0,sizeof(map)); for(i = 1; i <= r; i ++) scanf("%s",str[i]+1); map[1][start] = 1; i = 1; j = start; flag = -2; while(1) { if(str[i][j] == 'N') { if(i-1 < 1)//走出地图，flag标记为-1 flag = -1; else if(map[i-1][j] == 0)//步数为0，说明没有访问过该处 { map[i-1][j] = map[i][j] + 1;//步数增加 --i;//位置也要改变 } else //以上条件都不满足，说明形成环状 flag = map[i-1][j];//记录环开始的地方的步数 } else if(str[i][j] == 'S')//同上 { if(i + 1 > r) flag = -1; else if(map[i+1][j] == 0) { map[i+1][j] = map[i][j] + 1; ++i; } else flag = map[i+1][j]; } else if(str[i][j] == 'W')//同上 { if(j-1 < 1) flag = -1; else if(map[i][j-1] == 0) { map[i][j-1] = map[i][j] + 1; j--; } else flag = map[i][j-1]; } else if(str[i][j] == 'E')//同上 { if(j + 1 > c) flag = -1; else if( map[i][j+1] == 0) { map[i][j+1] = map[i][j] + 1; ++j; } else flag = map[i][j+1]; } if(flag != -2)//如果不为初始值，说明已经走出地图或者形成环 break; } if(flag == -1) printf("%d step(s) to exit ",map[i][j]); else printf("%d step(s) before a loop of %d step(s) ",flag-1,map[i][j]-flag+1); } return 0; }
这几天补搜索题实在是费脑子