HDOJ 1312 Red and Black

最新推荐文章于 2020-04-22 22:56:52 发布

moye_chen

最新推荐文章于 2020-04-22 22:56:52 发布

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分类专栏： BFS

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本文介绍了一道经典的迷宫寻路问题HDOJ1312，通过宽度优先搜索(BFS)算法来计算从起始位置可以到达的所有黑色瓷砖数量。输入包括地图尺寸及布局，输出可达的黑色瓷砖总数。

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HDOJ 1312 Red and Black

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

入门题,看了学长给的PPT,自己想了想ac了

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int maxsize=200;
struct point
{
    int x,y;
};
queue <point> q;
int flag[maxsize][maxsize];
char mapp[maxsize][maxsize];
int a[]= {0,0,1,-1};
int b[]= {1,-1,0,0};
int count1=1;
void bfs(point p);
int n,m;
int sx,sy;
int main()
{
    while(cin>>m>>n)
    {
        if(n==0)break;
        memset(flag,0,sizeof(flag));
        count1=1;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                cin>>mapp[i][j];
                if(mapp[i][j]=='@')
                {
                    sx=i;
                    sy=j;
                }
            }
        }
        point s= {sx,sy};
        bfs(s);

        cout<<count1<<endl;
    }


    return 0;
}
void bfs(point p)
{
    q.push(p);
    while(q.empty()==0)
    {
        point x=q.front();
        q.pop();
        for(int k=0; k<4; k++)
        {
            int i=x.x+a[k];
            int j=x.y+b[k];
            if(i>=0&&i<n&&j>=0&&j<m&&mapp[i][j]=='.'&&flag[i][j]==0)
            {
                flag[i][j]=1;
                point tmp= {i,j};
                q.push(tmp);
                count1++;
            }
        }
    }
}