poj 1573(搜索)

Robot Motion

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12351		Accepted: 5982

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 0

Sample Output

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

Source

Mid-Central USA 1999

 

用vis 数组做计数器。如果碰到已标记的就证明走回来了。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iostream>
using namespace std;
char graph[][];
int n,m,k;
int vis[][];
struct Node
{
    int x,y;
    int step;
} s;
bool check(int x,int y)
{
    if(x<||x>=n||y<||y>=m) return false;
    return true;
}
void bfs()
{
    memset(vis,,sizeof(vis));
    Node s;
    s.x = ,s.y = k-,s.step=;
    queue<Node> q;
    q.push(s);
    vis[s.x][s.y] = ;
    while(!q.empty())
    {
        Node now = q.front();
        q.pop();
        Node next;
        if(graph[now.x][now.y]=='W')
        {
            next.x = now.x;
            next.y = now.y-;
        }
        if(graph[now.x][now.y]=='S')
        {
            next.x = now.x+;
            next.y = now.y;
        }
        if(graph[now.x][now.y]=='E')
        {
            next.x = now.x;
            next.y = now.y+;
        }
        if(graph[now.x][now.y]=='N')
        {
            next.x = now.x-;
            next.y = now.y;
        }
        next.step = now.step+;
        if(check(next.x,next.y))
        {
            if(vis[next.x][next.y])  /// 如果被访问过了,则进入了循环
            {
                printf("%d step(s) before a loop of %d step(s)\n",vis[next.x][next.y]-,next.step-vis[next.x][next.y]);
                return ;
            }
            else
            {
                vis[next.x][next.y] = next.step;
                q.push(next);
            }
        }
        else
        {
            printf("%d step(s) to exit\n",next.step-);
            return;
        }

    }
    return;
}

int main()
{
    int t = ;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF&&n+m+k)
    {
        for(int i=; i<n; i++){
            scanf("%s",graph[i]);
        }
        bfs();
    }
    return ;
}

还写了个DFS的。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iostream>
using namespace std;
int graph[][];
int n,m,k;
int vis[][];
struct Node
{
    int x,y;
    int step;
}s;
bool check(int x,int y)
{
    if(x<||x>=n||y<||y>=m) return false;
    return true;
}
void dfs(int x,int y,int cnt){
    vis[x][y] = cnt;
    int nextx,nexty,step;
    if(graph[x][y]==){
        nextx = x;
        nexty = y - ;
    }
    if(graph[x][y]==){
        nextx = x+;
        nexty = y;
    }
    if(graph[x][y]==){
        nextx = x;
        nexty = y + ;
    }
    if(graph[x][y]==){
        nextx = x-;
        nexty = y;
    }
    step = cnt+;
    if(check(nextx,nexty)){
        if(vis[nextx][nexty]){
            printf("%d step(s) before a loop of %d step(s)\n",vis[nextx][nexty]-,step-vis[nextx][nexty]);
            return;
        }else{
            dfs(nextx,nexty,step);
        }
    }else{
        printf("%d step(s) to exit\n",step-);
    }
}
int main()
{
    int t = ;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF&&n+m+k)
    {
        char s[];
        for(int i=; i<n; i++){
            scanf("%s",s);
            for(int j=;j<m;j++){
                if(s[j]=='W') graph[i][j]=;
                if(s[j]=='S') graph[i][j]=;
                if(s[j]=='E') graph[i][j]=;
                if(s[j]=='N') graph[i][j]=;
            }
        }
        memset(vis,,sizeof(vis));
        dfs(,k-,);
    }
    return ;
}