DFS+BFS练习

A - Red and Black

 https://vjudge.net/contest/241948#problem/A

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

BFS代码实现：

#include 
#include 
#include 
using namespace std;
int w,h;
char a[100][100];
int d[4][2]={0,1,0,-1,1,0,-1,0};
int vis[100][100];
struct node{
	int x,y;
};
void BFS(int sx,int sy)
{
	queue q;
	node now,next;
	memset(vis,0,sizeof(vis));
	now.x=sx;
	now.y=sy;
	q.push(now);
	int ans=1;
	vis[sx][sy]=1;
	while(!q.empty())
	{

		now=q.front();
		q.pop();
	for(int i=0;i<4;i++)
	{
		next.x=now.x+d[i][0];
		next.y=now.y+d[i][1];
		if(next.x>=0&&next.x=0&&next.y

DFS代码实现：

一直有问题，找了非常长时间的bug!!!，DFS里if的条件应该是'.'或者'@',因为第一个总是从@开始的。要么就直接写成不等于'#'也成

#include 
#include 
using namespace std;
int w,h,ans;
char a[100][100];
int d[4][2]={0,1,0,-1,1,0,-1,0};
int vis[100][100];
void DFS(int sx,int sy){
    	if(sx>=0&&sx=0&&sy