POJ-1979 Red and Black （DFS or BFS 暴力）

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字数统计: 552 | 阅读时长 ≈ 3

描述

传送门：POJ-1979 Red and Black

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.

输入描述

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.

输出描述

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).

示例

输入

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20
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6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

输出

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3
4
45
59
6
13

题解

题目大意

给定一个W*H的矩形，里面的值为“.” “#”，其中“.”代表可到达的，“#”表示障碍，某人在一个“@”的起始点，求他所能到达的格子有多少个（包括第一所占的格子）。

思路

深搜广搜都行，直接暴力就好。

代码

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#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
const int MAXN = 1e3 +10;
using namespace std;
int dx[5] = {1, -1, 0, 0};
int dy[5] = {0, 0, 1, -1};
char MAP[MAXN][MAXN];
int book[MAXN][MAXN];
int w, h, res, x, y;

void dfs(int x, int y){
    if(!book[x][y] && MAP[x][y] != '#'){
        book[x][y] = 1;
        res++;
        for(int i = 0; i < 4; i++){
            int xx = x, yy = y;
            xx += dx[i]; yy += dy[i];
            if(xx >= 0 && yy >= 0 && xx < h && yy < w){
                dfs(xx, yy);
            }
        }
    }
    else return;
}

int main(){
    while(cin >> w >> h){
        if(w == 0 && h == 0) break;
        memset(book, 0, sizeof(book));
        for(int i = 0; i < h; i++)
            scanf("%s", MAP[i]);
        for(int i = 0; i < h; i++)
            for(int j = 0; j < w; j++)
                if(MAP[i][j] == '@') x = i, y = j;
        res = 0;
        dfs(x, y);
        cout << res << endl;
    }
}