POJ 1573 Robot Motion【是搜索，就不要纠结是DFS还是BFS】

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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

Output

Sample Input

3 6 5
NEESWE
WWWESS
SNWWWW
4 5 1
SESWE
EESNW
NWEEN
EWSEN
0 0 0

Sample Output

10 step(s) to exit
3 step(s) before a loop of 8 step(s)

Source

原题链接：http://poj.org/problem?id=1573

题意：有一个N*M的区域，机器人从第一行的第几列进入，该区域全部由‘N‘ , ‘S‘ , ‘W‘ , ‘E‘ ，走到某个区域的时候只能按照该区域指定的方向进行下一步，问你机器人能否走出该片区域，若不能，输入开始绕圈的步数和圈的大小。

DFS还是BFS对于这题有影响吗？

对，有，BFS 0ms！！DFS 16 ms！！！

搜索我是渣渣，看两份到吗的数据，POJ。

BFS AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
struct node
{
    int x,y;
    node(int x=0,int y=0):x(x),y(y) {};
    //装X的写法
};
char a[15][15];
bool vis[15][15];
int step[15][15];
int n,m,s;
void BFS()
{
    queue<node>q;
    memset(vis,false,sizeof(vis));
    step[0][s-1]=0;
    q.push(node(0,s-1));
    int stepp=-1;
    while(!q.empty())
    {
        node now=q.front();
        q.pop();
        stepp++;
        if(now.x<0||now.x>=n||now.y<0||now.y>=m)
        {
            printf("%d step(s) to exit\n",stepp);
            return ;
        }
        if(vis[now.x][now.y])
        {
            printf("%d step(s) before a loop of %d step(s)\n",step[now.x][now.y],stepp-step[now.x][now.y]);
            return ;
        }
        vis[now.x][now.y]=true;
        step[now.x][now.y]=stepp;
        if(a[now.x][now.y]=='S')
        {
            q.push(node(now.x+1,now.y));
        }
        else if(a[now.x][now.y]=='N')
        {
            q.push(node(now.x-1,now.y));
        }
        else if(a[now.x][now.y]=='W')
        {
            q.push(node(now.x,now.y-1));
        }
        else if(a[now.x][now.y]=='E')
        {
            q.push(node(now.x,now.y+1));
        }
    }
}
int main()
{
    while(cin>>n>>m>>s,n,m,s)
    {
        for(int i=0; i<n; i++)
            cin>>a[i];
        BFS();
    }
    return 0;
}

/**
3 3 1
SWW
SSN
EEN
*/

DFS AC代码：15ms

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char a[15][15];
bool vis[15][15];
int step[15][15];
int n,m,s,stepp;
void DFS(int x,int y,int t)
{
    stepp++;
    if(x<0||x>=n||y<0||y>=m)
    {
        printf("%d step(s) to exit\n",stepp);
        return ;
    }
    if(vis[x][y])
    {
        printf("%d step(s) before a loop of %d step(s)\n",step[x][y],stepp-step[x][y]);
        return ;
    }
    vis[x][y]=true;
    step[x][y]=stepp;
    int xx,yy;
    if(a[x][y]=='S')
    {
        //x+=1;
        xx=x+1;
        yy=y;
    }
    else if(a[x][y]=='N')
    {
        //x-=1
        xx=x-1;
        yy=y;
    }
    else if(a[x][y]=='W')
    {
        //y-=1;
        xx=x;
        yy=y-1;
    }
    else if(a[x][y]=='E')
    {
        //y+=1;
        xx=x;
        yy=y+1;
    }
    DFS(xx,yy,stepp+1);
}
int main()
{
    while(cin>>n>>m>>s,n,m,s)
    {
        for(int i=0; i<n; i++)
            cin>>a[i];
        stepp=-1;
        memset(vis,false,sizeof(vis));
        DFS(0,s-1,0);
    }
    return 0;
}

POJ 1573 Robot Motion【是搜索，就不要纠结是DFS还是BFS】

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原文地址：http://blog.csdn.net/hurmishine/article/details/52150066