HDU 1312 Red and Black（最简单也是最经典的搜索）

时间：2023-01-14 20:09:44

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http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25397    Accepted Submission(s): 15306

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
Source
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分析：
只能上下左右四个方向走，问你可以走的块最多是多少？#不能走
小技巧：走过的地方字符就变为#
先用dfs写一下，有时间再用bfs写
code：
#include<bits/stdc++.h>

using namespace std;

#define max_v 25

char G[max_v][max_v];

int n,m;

int sx,sy;

int step;

int dir[][]={,,,,,-,-,};

void dfs(int x,int y)

{

    int xx,yy;

    for(int i=;i<;i++)

    {

        xx=x+dir[i][];

        yy=y+dir[i][];

        if(xx>=&&xx<n&&yy>=&&yy<m&&G[xx][yy]!='#')

        {

            step++;

            G[xx][yy]='#';

            dfs(xx,yy);

        }

    }

}

int main()

{

    while(~scanf("%d %d",&m,&n))

    {

        if(n==&&m==)

            break;

        getchar();

        for(int i=;i<n;i++)

        {

            for(int j=;j<m;j++)

            {

                cin>>G[i][j];

                if(G[i][j]=='@')

                {

                    sx=i;

                    sy=j;

                }

            }

        }

        step=;

        G[sx][sy]='#';

        dfs(sx,sy);

        cout<<step<<endl;

    }

    return ;

}

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