noj1010 基本的迷宫问题

最新推荐文章于 2022-11-13 20:00:35 发布

原创最新推荐文章于 2022-11-13 20:00:35 发布 · 487 阅读

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CC 4.0 BY-SA版权

本文介绍了一个迷宫逃脱问题的解决方案，使用广度优先搜索（BFS）算法寻找从指定起点到迷宫出口的路径，并通过回溯记录路径。该算法适用于方形迷宫，迷宫由0（通道）和1（墙壁）组成。

The Inescapable Maze

Time Limit: 1000ms Memory Limit: 65536K

Total Submissions: 362 Accepted: 138

Description

Jack and his friends were trapped in a maze of amusement park. Please help them to find the right way to escape.

This square maze is reperented as 0,1 text. 0 is a way, 1 is a wall.

The size of maze is N. The entrance at (1,1), the exit at (N,N). You can only escape from exit.

Input

The first line contain three integers, N, x, y. (N<=80)

N is the size of maze. x, y is the Jack and friends's current location.

Your task is to find the way from (x,y) to (N,N)

Output

If you find the way out, print 'Found' then following with the maze model with '#' to show the way.

If you could't find the way, print 'Not Found'.

Sample Input

6 5 2
0 0 1 1 1 1
1 0 0 0 0 1
1 1 1 0 1 1
1 0 0 0 1 1
1 0 1 0 1 1
1 0 1 0 0 0

Sample Output

Found
0 0 1 1 1 1
1 0 0 # # 1
1 1 1 # 1 1
1 0 0 # 1 1
1 0 1 # 1 1
1 0 1 # # #

#include <iostream>
#include<stdio.h>
#include<queue>
#include<stack>
#include<string.h>
using namespace std;

int t;
int m,n;

struct point
{
    int x;
    int y;
}buf;

int vis[100][100];
int mapp[1000][100];
int keep[100][100];
int dir[][2]={{1,0},{-1,0},{0,1},{0,-1}};
queue<point> q;
stack<point >s;
int flag;

void  bfs()
{
    flag=0;
    vis[buf.x][buf.y]=1;
    while(!q.empty())
    {
        for(int i=0;i<4;i++)
        {
            buf.x=q.front().x+dir[i][0];
            buf.y=q.front().y+dir[i][1];
            if(1<=buf.x&&buf.x<=t&&1<=buf.y&&buf.y<=t&& !vis[buf.x][buf.y] &&mapp[buf.x][buf.y]==0)
            {
                vis[buf.x][buf.y]=1;
                keep[buf.x][buf.y]=i;
                q.push(buf);
                if(buf.x==t&&buf.y==t)
                {
                    flag=1;
                    break;
                }
            }
        }
        q.pop();
        if(buf.x==t&&buf.y==t)
        {
            flag=1;
            break;
        }
    }
}

int main()
{
    memset(vis,0,sizeof(vis));
    scanf("%d%d%d",&t,&m,&n);
    for(int i=1;i<=t;i++)
        for(int j=1;j<=t;j++)
            scanf("%d",&mapp[i][j]);
        buf.x=n;
        buf.y=m;
        q.push(buf);
        bfs();
        if(flag==0)
            printf("Not Found\n");
        else
        {
            printf("Found\n");
        point nbuf={t,t};
        s.push(nbuf);
        while(buf.x!=n||buf.y!=m)
        {
            nbuf.x=buf.x-dir[keep[buf.x][buf.y]][0];
            nbuf.y=buf.y-dir[keep[buf.x][buf.y]][1];
            s.push(nbuf);
            buf.x=nbuf.x;
            buf.y=nbuf.y;
        }
        while(!s.empty())
        {
            mapp[s.top().x][s.top().y]=2;
            s.pop();
        }
       for(int i=1;i<=t;i++)
        for(int j=1;j<=t;j++)
        {
            if(mapp[i][j]==0||mapp[i][j]==1)
            printf("%d",mapp[i][j]);
            else
                printf("#");
            if(j!=t)
                printf(" ");
           if(j==t&&i!=t)
            printf("\n");
        }
        }
        return 0;
}