zoukankan      html  css  js  c++  java
  • hdoj_1312Red and Black

    Red and Black

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 5175    Accepted Submission(s): 3370


    Problem Description
    There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

    Write a program to count the number of black tiles which he can reach by repeating the moves described above. 
     

    Input
    The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

    There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

    '.' - a black tile 
    '#' - a red tile 
    '@' - a man on a black tile(appears exactly once in a data set) 
     

    Output
    For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself). 
     

    Sample Input
    6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
     

    Sample Output
    45 59 6 13

    #include <iostream>
    #include <cstring>
    using namespace std;
    #pragma warning(disable : 4996)
    #define MAX 25
    char maps[MAX][MAX];
    int w, h, ans;
    const int moves[4][2] = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
    
    void dfs(int x, int y)
    {
    	maps[x][y] = '#';
    	for(int i = 0; i < 4; i++)
    	{
    		int p = x + moves[i][0];
    		int q = y + moves[i][1];
    		if(p >= 1 && p <= h && q >= 1 && q <= w && maps[p][q] == '.')
    		{
    			maps[p][q] = '#';
    			ans++;
    			dfs(p, q);
    		}
    	}
    
    }
    
    int main()
    {
    	freopen("in.txt", "r", stdin);
    	int x, y;
    	while (cin >>w >> h)
    	{
    		if(w == 0 && h == 0)
    		{
    			break;
    		}
    		for(int i = 1; i <= h; i++)
    		{
    			for(int j = 1; j <= w; j++)
    			{
    				cin >> maps[i][j];
    				if(maps[i][j] == '@')
    				{
    					x = i;
    					y = j;
    				}
    			}
    		}
    		ans = 1;
    		dfs(x, y);
    		cout << ans << endl;
    	}
    	return 0;
    }



  • 相关阅读:
    最长公共子序列
    学习MySQL常用操作命令
    using的几种用法
    C++循环的简单使用【闲来写来练练手~】
    使用【数据库收缩功能】实现多个数据文件的合并
    Google的十个核心技术(摘自CSDN)
    OPENGL入门学习
    dive into python 第 5 章 对象和面向对象
    [转]已知两圆圆心坐标及半径求两圆交点 (C语言|参数方程求解)
    The Python Tutorial 4. More Control Flow Tools的一些小记
  • 原文地址:https://www.cnblogs.com/lgh1992314/p/5835122.html
Copyright © 2011-2022 走看看