HDU 1312Red and Black(简单搜索 bfs或dfs) - c++编程基础

Red and Black

Time Limit: 2000/1000 MS ( Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6908 Accepted Submission(s): 4387

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

Source

Asia 2004, Ehime (Japan), Japan Domestic

题目意思就不多说了，很容易理解，@为起点，可以上下左右四个方向扩展，

只能走点，问可以到达多少个位置。

DFS AC代码：

#include  
#include  
#include  
#include  
using namespace std;  
int r,c;  
int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}};  
char s[25][25];  
int visi[25][25];  
  
void dfs(int a,int b)  
{  
    int i;  
    for(i=0;i<4;i++)  
    {  
        int ta,tb;  
        ta=a+dir[i][0],tb=b+dir[i][1];  
        if(ta>=0&&ta=0&&tb

//0MS

BFS AC代码：

#include  
#include  
#include  
#include  
#include  
using namespace std;  
int r,c;  
int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}};  
char s[25][25];  
int visi[25][25];  
  
void bfs(int t)  
{  
    queue mq;  
    mq.push(t);  
    while(!mq.empty())  
    {  
        int p=mq.front();  
        mq.pop();  
        int ta=p/c,tb=p%c;  
        for(int i=0;i<4;i++)  
        {  
            int tta=dir[i][0]+ta,ttb=dir[i][1]+tb;  
            if(tta>=0&&tta=0&&ttb