Red and Black

转载：https://blog.youkuaiyun.com/hurmishine/article/details/50927317

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

基本上都用的是DFS，但我的第一直觉用的是BFS，但是不知道如何结束，后来发现一般DFS都是要用到递归，所以要出现return来实现递归时条件不满足的情况，比如说在迷宫中找人，找终点，就是当前访问的坐标等于要找的人或终点的坐标。然而BFS一般都是用队列实现，while(队列非空)来判断，因为把所有的可能的坐标都放在队列里，然后出队入队直到把所有值都找遍了，就无值入队，队列为空时结束。

这里没有设visited数组，是因为本身a数组就可以判断，判断的办法即把所有访问过的置为’#’

用BFS

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>

using namespace std;
char a[21][21];
int vis[21][21];
int w, h;//w是列，h是行
int dir[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };
struct position
{
    int x;
    int y;
};
int sum;
bool judge(int x, int y)
{
    if (x > 0 && x <= h&&y>0 && y <= w)
        return true;
    return false;
}
void Bfs(position place)
{
    queue<position> q;
    q.push(place);

    position pos;
    position temppos;
    while (!q.empty())
    {
        pos = q.front();
        q.pop();
        for (int i = 0; i < 4; i++)
        {
            temppos.x = pos.x + dir[i][0];
            temppos.y = pos.y + dir[i][1];
            if (judge(temppos.x,temppos.y) &&a[temppos.x][temppos.y]=='.')
            {
                a[temppos.x][temppos.y] = '#';
                q.push(temppos);
                sum++;
            }
        }
    }
}
int main()
{
    while (scanf_s("%d%d", &w, &h) && w != 0 && h != 0)
    {
        memset(a, 0, sizeof(a));
        memset(vis, 0, sizeof(vis));
        position begin;
        for (int i = 1; i <= h; i++)
        {
            getchar();
            for (int j = 1; j <= w; j++)
            {
                a[i][j] = getchar();
                if (a[i][j] == '@')
                {
                    begin.x = i;
                    begin.y = j;
                }
            }
        }
        sum = 1;
        Bfs(begin);
        printf("%d\n", sum);
    }
    return 0;
}

用DFS

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>

using namespace std;
char a[21][21];
int w, h;//w是列，h是行
int dir[4][2] = { { -1,0 },{ 1,0 },{ 0,-1 },{ 0,1 } };
struct position
{
    int x;
    int y;
};
int sum;
void Dfs(position place)
{
    if (place.x <= 0 || place.x > h || place.y <= 0 || place.y > w||a[place.x][place.y]=='#')
        return;
    a[place.x][place.y] = '#';
    sum++;
    for (int i = 0; i < 4; i++)
    {
        position temp;
        temp.x = place.x + dir[i][0];
        temp.y = place.y + dir[i][1];
        Dfs(temp);
    }
}
int main()
{
    while (scanf_s("%d%d", &w, &h) && w != 0 && h != 0)
    {
        memset(a, 0, sizeof(a));
        position begin;
        for (int i = 1; i <= h; i++)
        {
            getchar();
            for (int j = 1; j <= w; j++)
            {
                a[i][j] = getchar();
                if (a[i][j] == '@')
                {
                    begin.x = i;
                    begin.y = j;
                }
            }
        }
        sum = 0;
        Dfs(begin);
        printf("%d\n", sum);
    }
    return 0;
}