文章标题 HDU 1312 : Numerically Speaking （BFS）

最新推荐文章于 2021-08-02 07:57:33 发布

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本文介绍了一个基于广度优先搜索(BFS)算法解决迷宫寻路问题的方法。在一个由红黑方格组成的矩形迷宫中，从指定的起点出发，计算能够到达的所有黑色方格数量。通过使用C++实现的BFS算法，有效解决了这一问题。

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Numerically Speaking

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

题意：给我们一个矩阵，里面有红色的（用 . 表示）和黑色的（用#表示），有一个人站在一个红色的点上面（用@表示），现要求出这个人能够走几个红色的点，这个人遇到红色的点可以走，遇到黑色的点就不能走。
分析：用BFS从人的位置（@的位置）开始遍历，遍历包含@这个点的连通快的数目就是此题所要的答案。
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
char mp[25][25];
int w,h; 
int dx[4]={1,0,-1,0};//四个方向 
int dy[4]={0,1,0,-1};
struct node {
    int x,y;
};
int bfs(int x,int y){
    node t;//记录当前这个点 
    t.x=x;t.y=y;
    queue<node>q;
    q.push(t);//放进队列 
    int cnt=1;
    mp[x][y]='#';//标记已经走过了 
    while (!q.empty()){
        node tmp=q.front();q.pop();
        int nx,ny;
        for (int i=0;i<4;i++){//遍历四个方向 
            nx=tmp.x+dx[i];
            ny=tmp.y+dy[i];
            if (nx>=0&&nx<h&&ny>=0&&ny<w&&mp[nx][ny]=='.'){//看是否符合 
                cnt++;//符合的话计数+1 
                node tt;tt.x=nx;tt.y=ny;
                q.push(tt);//放进队列 
                mp[nx][ny]='#';//标记已经走过 
            }
        }
    }
    return cnt;
}
int main ()
{
    int x,y;
    while (cin>>w>>h){
        if (w==0&&h==0) break;
        for (int i=0;i<h;i++){
            for (int j=0;j<w;j++){
                cin>>mp[i][j];
                if (mp[i][j]=='@'){//找出人的位置 
                    x=i;
                    y=j;
                }
            }
        }
        int ans=bfs(x,y);//得到答案 
        printf ("%d\n",ans);
    }   
    return 0;
}