Red and Black

最新推荐文章于 2020-11-01 16:53:15 发布

huxiansheng__

最新推荐文章于 2020-11-01 16:53:15 发布

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分类专栏： HDU

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本文介绍了一种通过简单搜索算法解决的问题，即在一个由红色和黑色瓷砖组成的矩形房间中，计算从初始位置可达的黑色瓷砖数量。文章提供了一个详细的算法实现过程，包括输入输出示例。

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Red and Black

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 7 Accepted Submission(s) : 4

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

题解：这个题就是用一个简单的搜索，从起点开始进队列，只要是可以走的点都进队列，然后标记为不可走的。如果队列里的点无法再进行向四个方向走就出队列，只要有进入队列的就是可以走的点。最后所有的点出队列。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
struct node
{
    int x,y;
};
char cha[25][25];
int dir[4][2]={0,1,1,0,-1,0,0,-1};
int main()
{
    int n,m,i,j,sx,sy,c;
    while(scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)break;
        c=1;
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=n;j++)
            {
                cin>>cha[i][j];
                if(cha[i][j]=='@')
                {
                    sx=i;
                    sy=j;
                }
            }
        }
        node d1;
        d1.x=sx;
        d1.y=sy;
        cha[sx][sy]=='#';
        queue<node>q;
        q.push(d1);
        while(!q.empty())
        {
            node d2;
            d2=q.front();
            q.pop();
        for(i=0;i<4;i++)
        {
            node d3;
            d3.x=d2.x+dir[i][0];
            d3.y=d2.y+dir[i][1];
            if(cha[d3.x][d3.y]=='.'&&d3.x>0&&d3.x<=m&&d3.y>0&&d3.y<=n)
            {
                c++;
                cha[d3.x][d3.y]='#';
                q.push(d3);
            }

        }
        }
        cout<<c<<endl;

    }
}