HDU 1312：Red and Black（BFS）

Red and Black

Time limit:1000 ms Memory limit:32768 kB OS:Windows

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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

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题意：

以’@’符号为起点，’.’可以走，’#’不能走，方向为上下左右，问最多可以走多少格子。

解题思路：

这么水的。。直接BFS就好了，注意下H和W的次序就好.

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Code：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

const int maxn=20+5;
int vis[maxn][maxn];
char mp[maxn][maxn];
int n,m;
int dx[]= {0,1,0,-1};
int dy[]= {1,0,-1,0};

struct Node
{
    int x,y;
};
queue<Node> Q;

bool check(int x,int y)
{
    if(x<0||x>=m||y<0||y>=n)
        return false;
    if(vis[x][y])
        return false;
    if(mp[x][y]!='.')
        return false;
    return true;
}

int BFS(int x1,int y1)
{
    vis[x1][y1]=1;
    int ans=0;
    Q.push((Node){x1,y1});
    while(!Q.empty())
    {
        ans++;
        Node &u=Q.front();
        Q.pop();
        for(int i=0; i<4; i++)
        {
            int x=u.x+dx[i];
            int y=u.y+dy[i];
            if(check(x,y))
            {
                vis[x][y]=1;
                Q.push(Node{x,y});
            }
        }
    }
    return ans;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(m==0&&n==0)
            break;
        for(int i=0; i<m; i++)
            scanf("%s",mp[i]);
        mem(vis,0);
        for(int i=0; i<m; i++)
            for(int j=0; j<n; j++)
                if(mp[i][j]=='@')
                {
                    printf("%d\n",BFS(i,j));
                    break;
                }
    }
    return 0;
}