POJ1979/HDU 1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8336    Accepted Submission(s): 5184

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
  
  

 

题解：一道经典的搜索题。

DFS：

/*
题意：'#'不可走，'.'可走，'@'为起点。
      求从'@'开始走，一共可以走完多少个'.','@'也算一个。
思路：dfs深度遍历周围的四个方向
优化：遍历前，图的四周加边。
注意：先输入高度，再输入宽。也就是先输入行数再输入列数。

*/
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

char map[25][25];
int w, h;
int sx, sy; //起点
int ans;

void dfs(int x, int y) //深度遍历
{
    if(map[x][y] == '#') return; //不可走，马上跳出
    ans++;   //可以走 结果+1
    map[x][y] = '#'; // 标记已走
    dfs(x-1, y); dfs(x+1, y); //遍历周围的四个方向
    dfs(x, y-1); dfs(x, y+1);
}
int main()
{
    while(scanf("%d%d", &w, &h) != EOF) //先输入列数 再输入行数
    {
        if(w == 0) break; // 输入0即结束
        for(int i = 1; i <= h; i++)
        {
            for(int j = 1; j <= w; j++)
            {
                cin>>map[i][j];
                if(map[i][j] == '@') //起点
                {
                    sx = i;
                    sy = j;
                }
            }
        }

        for(int i = 0; i <= h+1; i++) map[i][0] = map[i][w+1] = '#'; //第0列和第w+1列加边 不可走
        for(int i = 0; i <= w+1; i++) map[0][i] = map[h+1][i] = '#'; //第0行和第h+1行加边 不可走
        ans = 0; //初始化结果
        dfs(sx, sy);
        printf("%d\n", ans);
    }
    return 0;
}

BFS:

/************************************************************************/
/*
  poj 1979 red and black
使用方法为BFS                                                                      */
/************************************************************************/
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
int n,m,sx,sy,step;
struct Node{
	int x,y;
};
#define MAX 25

char map[MAX][MAX];
int vist[MAX][MAX];
int dist[4][2]={1,0,-1,0,0,1,0,-1};

int inMAP(int x,int y){
	if(x<0 || x>=n || y<0 || y>=m)
		 return 0;
   return 1;
}
void BFS(){
    queue <Node> Q;
	Node start,now,next;
	start.x = sx;
	start.y = sy;
	step=1;//开始位置
	vist[sx][sy]=1;
	Q.push(start);
	while(!Q.empty()){
		now = Q.front();
		Q.pop();
		for(int i=0;i<4;i++)
		{
		
			next.x =now.x +dist[i][0];
			next.y =now.y +dist[i][1];
			//满足条件的情况
			if(!vist[next.x][next.y] && map[next.x][next.y]=='.' && inMAP(next.x,next.y)){
			   	vist[next.x][next.y]=1;
                  step+=1;
				 Q.push(next);
			}
		}
	}

}
int main(){

	int i,j;
	//注意这里的行和列，被坑了。
	while(cin>>m>>n){     //m=6列 y ，行n=9 x
		memset(vist,0,sizeof(vist));
		if(n+m==0)break;
        for(i=0; i<n;i++)
		  for(j=0;j<m;j++)
			  cin>>map[i][j];
		  for(i=0;i<n;i++)
			  for(j=0;j<m;j++)
				  if(map[i][j]=='@')
				  {
					  sx=i;
					  sy=j;
					  break;
				  }
        
		       BFS();
			  cout<<step<<endl;
	}
	return 0;
}

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