zoukankan      html  css  js  c++  java
  • POJ 2157 Maze

    Maze
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 3183   Accepted: 996

    Description

    Acm, a treasure-explorer, is exploring again. This time he is in a special maze, in which there are some doors (at most 5 doors, represented by 'A', 'B', 'C', 'D', 'E' respectively). In order to find the treasure, Acm may need to open doors. However, to open a door he needs to find all the door's keys (at least one) in the maze first. For example, if there are 3 keys of Door A, to open the door he should find all the 3 keys first (that's three 'a's which denote the keys of 'A' in the maze). Now make a program to tell Acm whether he can find the treasure or not. Notice that Acm can only go up, down, left and right in the maze.

    Input

    The input consists of multiple test cases. The first line of each test case contains two integers M and N (1 < N, M < 20), which denote the size of the maze. The next M lines give the maze layout, with each line containing N characters. A character is one of the following: 'X' (a block of wall, which the explorer cannot enter), '.' (an empty block), 'S' (the start point of Acm), 'G' (the position of treasure), 'A', 'B', 'C', 'D', 'E' (the doors), 'a', 'b', 'c', 'd', 'e' (the keys of the doors). The input is terminated with two 0's. This test case should not be processed.

    Output

    For each test case, in one line output "YES" if Acm can find the treasure, or "NO" otherwise.

    Sample Input

    4 4 
    S.X. 
    a.X. 
    ..XG 
    .... 
    3 4 
    S.Xa 
    .aXB 
    b.AG 
    0 0
    

    Sample Output

    YES 
    NO
    #include<stdio.h>
    #include<iostream>
    #include <string.h>
    #include <queue>
    using namespace std;
    typedef struct node
    {
        int x;
        int y;
        node(int a, int b)
        {
            x = a;
            y = b;
        }
        node()
        {
    
        }
    }Map;
    
    char Maze[25][25];
    int Dir[4][2] = {-1,0,1,0,0,-1,0,1};
    int key[10];
    
    void BFS(int sx, int sy, int m, int n)
    {
        queue<Map> Queue;
        Queue.push(Map(sx, sy));
        Map temp;
        Maze[sx][sy] = 'X';
        int Limit = 0;
        while(!Queue.empty() && Limit < 400)
        {
            ++Limit;
            temp = Queue.front();
            Queue.pop();
            if (Maze[temp.x][temp.y] >= 'A' && Maze[temp.x][temp.y] <= 'E')
            {
                if (key[Maze[temp.x][temp.y] - 'A'] == 0)
                {
                    Maze[temp.x][temp.y] = 'X';
                }
                else
                {
                    Queue.push(temp);
                    continue;
                }
            }
            for (int i = 0; i < 4; i++)
            {
                int x = temp.x + Dir[i][0];
                int y = temp.y + Dir[i][1];
                if (x >= 0 && x < m && y >= 0  && y < n && Maze[x][y] != 'X')
                {
                    if (Maze[x][y] == '.')
                    {
                        Maze[x][y] = 'X';
                        Queue.push(Map(x, y));
                    }
                    if (Maze[x][y] >= 'a' && Maze[x][y] <= 'e')
                    {
                        key[Maze[x][y] - 'a']--;
                        Maze[x][y] = 'X';
                        Queue.push(Map(x, y));
                    }
                    if (Maze[x][y] == 'G')
                    {
                        printf("YES
    ");
                        return;
                    }
                    if (Maze[x][y] >= 'A' && Maze[x][y] <= 'E')
                    {
                        Queue.push(Map(x, y));
                    }            
                }
            }
        }
        printf("NO
    ");
    }
    
    int main()
    {
        int m, n;
        int sx, sy;
        while(scanf("%d%d", &m, &n) != EOF)
        {
            if (m == 0 && n == 0)
            {
                break;
            }
            memset(key, 0, sizeof(key));
            for (int i = 0; i < m; i++)
            {
                scanf("%s", Maze[i]);
                for (int j = 0; j < n; j++)
                {
                    if (Maze[i][j] == 'S')
                    {
                        sx = i;
                        sy = j;
                    }
                    else
                    {
                        if (Maze[i][j] >= 'a' && Maze[i][j] <= 'e')
                        {
                            key[Maze[i][j] - 'a']++;
                        }
                    }
                }
            }
            BFS(sx, sy, m, n);
        }
        return 0;
    }
  • 相关阅读:
    gnuplot learn note
    command line text process
    raspberry pi boot without HDMI
    gnuplot运行方式
    读取外部excel文件
    DB2中Lob is closed. ERRORCODE=4470的解决
    Myeclipse项目编码
    Json使用
    数组元素全排列递归算法
    XmlHttpRequest IE 乱码问题
  • 原文地址:https://www.cnblogs.com/lzmfywz/p/3204303.html
Copyright © 2011-2022 走看看