ZOJ-2165（DFS）

最新推荐文章于 2020-07-24 18:26:18 发布

Asmiiiire

最新推荐文章于 2020-07-24 18:26:18 发布

阅读量356

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/qq_36539373/article/details/54561778

本文介绍了一种算法，用于解决在一个由红色和黑色瓷砖组成的矩形房间中，从一个初始位置出发，仅能沿黑色瓷砖移动的情况下，计算能够到达的所有黑色瓷砖数量的问题。文章通过深度优先搜索的方法实现了这一目标，并提供了完整的源代码。

Red and Black

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

Total Submission(s) : 67 Accepted Submission(s) : 49

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

Source

Asia 2004, Ehime (Japan), Japan Domestic

连通块问题，

当初怎么改都没改出来，

突然发现是先列后行。。

思路与最大黑区域相同：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

int Map[21][21];
char s[21][21];
int n,m;
int ax,ay;
int cnt;
int dir[4][2]={1,0,-1,0,0,1,0,-1};
int mx;
int my;

void dfs(int x,int y)
{
       for(int i=0;i<4;i++)
       {
           mx=x+dir[i][0];
           my=y+dir[i][1];
           if(Map[mx][my]==1&&mx>=0&&my>=0&&mx<n&&my<m)
           {
               cnt++;
               Map[mx][my]=0;
               dfs(mx,my);
           }
       }
}



int main()
{
    while(~scanf("%d%d",&m,&n)&&n&&m)
    {
        cnt=1;
        memset(Map,0,sizeof(Map));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                cin>>s[i][j];
                if(s[i][j]=='@')
                {
                    ax=i;
                    ay=j;
                }
                if(s[i][j]=='.')
                {
                    Map[i][j]=1;
                }
            }
        }
        Map[ax][ay]=0;
        dfs(ax,ay);
        cout<<cnt<<endl;
    }
    return 0;
}