• DFS入门__poj1979


    Red and Black
    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 26944   Accepted: 14637

    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) 
    The end of the input is indicated by a line consisting of two zeros. 

    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
    /*题目大意:在一个矩形房间里,里面填满着红砖与黑砖,每次走动只能走黑砖位置,而不能走红砖,
     *			可以上下左右四个方向走动,现在初始位置在黑砖位置上,试问从此黑砖位置开始能够到达的砖的数量是多少
     *算法分析:从起始位置开始四个方向进行dfs,若遇到红砖位置则将此砖块换为黑砖。换砖的次数即为到达的黑砖数量 
    
    */
    
    #include <iostream>
    #include <cstdio>
    using namespace std;
    
    char a[25][25];
    int n, m;
    int dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};				//四个方向数组 
    
    void dfs(int x, int y, int &res) {
    	a[x][y] = '#';
    	res ++ ;
    	for (int i = 0; i<4; i++) {
    		int nx = x + dir[i][0];
    		int ny = y + dir[i][1];
    		if (nx>=0 && nx<n && ny>=0 && ny<m && a[nx][ny] == '.')
    			dfs(nx, ny, res);
    	}
    }
    
    int main() {
    	
    	while (cin >> m >> n && (n+m)) {
    		
    		memset(a, 0, sizeof(a));
    		for (int i = 0; i<n; i++) {
    			for (int j = 0; j<m; j++) {
    				cin >> a[i][j];
    			}
    		}
    		int res = 0;
    		for (int i = 0; i<n; i++) {
    			for (int j = 0; j<m; j++) {
    				if (a[i][j] == '@') 
    					dfs(i, j, res);
    			}
    		}
    		cout << res << endl;
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/Tovi/p/6194826.html
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