Red and Black（poj-1979）

Posted 2020-09-27 雪晨

原题目：

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13


题目大意：
给出一个w列和h行的方阵，方阵中有‘.‘,‘#‘和‘@‘。其中‘.‘是可以移动的地方，‘#‘是红色的砖，是不可以移动的地方，‘@‘是最初人的地点就（图的入口），人可以上下左右移动，以0 0 为结束。
要求输出人可以移动的砖的个数。该题利用搜索。
注意：
先输入w，然后输入h，但是w是列数，h是行数。
代码：

#include<iostream>
#include<cstring>
using namespace std;

void dfs(int x,int y);
int w,h,tmp=0;

char ma[100][100];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};

int main(){
	while(cin>>h>>w){
		if(w==0&&h==0)  break;
		memset(ma,‘0‘,sizeof(ma));
		int ax,ay;
		for(int i=0;i<w;i++){
			for(int j=0;j<h;j++){
				cin>>ma[i][j];
				if(ma[i][j]==‘@‘){
					ax=i;
					ay=j;
				}
			}
		}
		tmp=0;
		dfs(ax,ay);
		cout<<tmp<<endl;
	}
	return 0;
}

void dfs(int x,int y){
	ma[x][y]=‘#‘;
	tmp++;
	for(int i=0;i<4;i++){
		int nx=x+dx[i];
		int ny=y+dy[i];
		if(ma[nx][ny]==‘.‘&&nx>=0&&nx<w&&ny>=0&&ny<h&&ma[nx][ny]==‘.‘){
			dfs(nx,ny);
		}
	}
}

Select Code

#include<iostream>
#include<cstring>
using namespace std;

void dfs(int x,int y);
int w,h,tmp=0;

char ma[100][100];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};

int main(){
	while(cin>>h>>w){
		if(w==0&&h==0)  break;
		memset(ma,‘0‘,sizeof(ma));
		int ax,ay;
		for(int i=0;i<w;i++){
			for(int j=0;j<h;j++){
				cin>>ma[i][j];
				if(ma[i][j]==‘@‘){
					ax=i;
					ay=j;
				}
			}
		}
		tmp=0;
		dfs(ax,ay);
		cout<<tmp<<endl;
	}
	return 0;
}

void dfs(int x,int y){
	ma[x][y]=‘#‘;
	tmp++;
	for(int i=0;i<4;i++){
		int nx=x+dx[i];
		int ny=y+dy[i];
		if(ma[nx][ny]==‘.‘&&nx>=0&&nx<w&&ny>=0&&ny<h&&ma[nx][ny]==‘.‘){
			dfs(nx,ny);
		}
	}
}

poj上不支持bits/stdc++的头文件，很尴尬。