Hdoj 1312.Red and Black 题解

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

Source

Asia 2004, Ehime (Japan), Japan Domestic

平平无奇的一道简单bfs问题，只要每次广搜入队的时候都统计一次就好了，最后返回结果并输出

代码

#include<bits/stdc++.h>
using namespace std;
int a,b,c,t;
const int d[][2]={ {-1,0},{0,1},{1,0},{0,-1} };
struct node
{
    int x;
    int y;
}st,ed;
int n,m;
char maps[21][21];
bool judge(node x)
{
    if(x.x<=m && x.x>=1 && x.y<=n && x.y>=1 && maps[x.x][x.y]=='.')
        return true;
    return false;
}
int bfs(node st)
{
    queue<node> q;
    q.push(st);
    maps[st.x][st.y] = '#';
    node now,next;
    int t = 0;
    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(judge(next))
            {
                q.push(next);
                t++;
                maps[next.x][next.y] = '#';
            }
        }

    }
    return t+1;//起点也算
}

int main()
{
    while(cin>>n>>m)
    {
        if(n==0 && m==0) break;
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
            {
                cin >> maps[i][j];
                if(maps[i][j]=='@')
                {
                    st.x = i; st.y = j;
                }
            }
        int ans = bfs(st);
        cout << ans << endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/MartinLwx/p/9902686.html