HDU1035 Robot Motion（dfs）

题目链接

Robot Motion

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13841    Accepted Submission(s): 6462

 

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)

navigate

英[ˈnævɪgeɪt] 美[ˈnævɪˌɡet]

vt.驾驶; 航行于; 使通过; vi. 航行; 驾驶;

结果有两种情况，要么退出，要么陷入循环。

 如果能退出则直接输出步数；如果陷入循环，则记录下开始循环的行和列再进行一次搜索。

AC代码：

#include<iostream>
#include<sstream>
#include<algorithm>
#include<string>
#include<cstring>
#include<iomanip>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
using namespace std;
int r,c,n,si,sj;
char map[11][11];
bool visited[11][11];
bool flag=0;//是否成功退出
int step=0;
void dfs(int i,int j,int s)
{
    if(i<1||i>r||j<1||j>c)//可以退出
    {
        flag=1;step=s;return;
    }
    else if(visited[i][j])//陷入循环 
    {
        si=i;sj=j;step=s;return;//记录下开始陷入循环的行和列 
    }
    visited[i][j]=1;
    if(map[i][j]=='N') dfs(i-1,j,s+1);//向上走
    else if(map[i][j]=='S') dfs(i+1,j,s+1);//向下走
    else if(map[i][j]=='E') dfs(i,j+1,s+1);//向右走
    else if(map[i][j]=='W') dfs(i,j-1,s+1);//向下走 
}
int main()
{
    while(cin>>r>>c&&r!=0)
    {
        cin>>n;
        memset(visited,0,sizeof(visited));
        for(int i=1;i<=r;i++) cin>>map[i]+1;
        
        dfs(1,n,0);//从第一行的第n列开始搜索，步数开始为0 
        if(flag) cout<<step<<" step(s) to exit
";
        else 
        {
            int loop=step;
            step=0;
            memset(visited,0,sizeof(visited));
            dfs(si,sj,0);//再深搜一遍测出循环的步数step 
            cout<<loop-step<<" step(s) before a loop of "<<step<<" step(s)
";
        }
        flag=0;
        step=0;
    }
}