ZOJ 2165

Red and Black

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

什么时候++是问题所在

昨天看的题目

今天忘记了题目输出要求

WA了三次

#include <iostream>
#include <queue>
#include <string>
#include <cstring>
using namespace std;
const int maxw = 25;
bool visit[maxw][maxw];
char tiles[maxw][maxw];
int step[4][2]={{1,0},{0,1},{0,-1},{-1,0}};
int w,h;
class Node
{
public:
    int x,y;

    void init(int a,int b)
    {
       x=a;
       y=b;
    }
};

int bfs(Node begin)
{
    int num = 0;
    queue<Node> Q;
    Q.push(begin);
    memset(visit,false,sizeof(visit));
    visit[begin.x][begin.y]=true;
    num++;
    while(Q.empty()!=true)
    {
        Node a = Q.front();
        Q.pop();
        if(visit[a.x][a.y]==false)
        {
            num++;
            visit[a.x][a.y]=true;
        }
        for(int i =0;i<4;i++)
        {
            int x=a.x+step[i][0];
            int y=a.y+step[i][1];
            if(x>=0&&x<h&&y>=0&&y<w&&visit[x][y]==false&&tiles[x][y]=='.')
            {
                Node b;
                b.init(x,y);
                Q.push(b);
            }
        }
    }
    return num;
}
int main()
{
    Node begin;
    while(cin>>w>>h)
    {
        if(w+h==0)
            return 0;
        string s;
        for(int i=0;i<h;i++)
        {
            cin>>s;
            for(int j=0;j<w;j++)
            {
                char ch = s.at(j);
                if(ch=='@')
                {
                    begin.init(i,j);
                }
                tiles[i][j]=ch;
            }
        }
        cout<<bfs(begin)<<endl;
    }
    return 0;
}