hdu 1312 Red and Black

最新推荐文章于 2024-02-20 18:00:32 发布

my1995

最新推荐文章于 2024-02-20 18:00:32 发布

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分类专栏：深度优先搜索文章标签： ACM character 深搜

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本文介绍了一个简单的程序设计问题，即计算在一个由黑色和红色瓷砖组成的矩形房间中，从初始位置出发可以到达的所有黑色瓷砖的数量。通过深度优先搜索算法实现了这一功能。

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我写了这道题目蛮久的，谁知道是水题

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

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
char G[25][25];
bool vist[25][25];
int m,n,co;
int dir[4][2]={{0,-1},{0,1},{-1,0},{1,0}};
void dfs(int x,int y){
    co++;
    vist[x][y]=1;
    for(int i=0;i<4;i++){
        int x1=x+dir[i][0];
        int y1=y+dir[i][1];
        if(x1>=0 && x1<m && y1>=0 && y1<n){
            if(G[x1][y1]=='.'&&!vist[x1][y1])
            {
                dfs(x1,y1);
                vist[x1][y1]=1;
            }

        }
    }
}
int main()
{
    int i,j,a,b;
    while(cin>>n>>m) //n为列，m为行
    {
        if(n==0&&m==0)
            break;
        memset(vist,0,sizeof(vist));
        for(i=0;i<m;i++)
        {
            cin>>G[i];
        }
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                if(G[i][j]=='@')
                {
                    a=i;
                    b=j;
                }
            }
        }
        co=0;
        dfs(a,b);
        cout<<co<<endl;
    }
    return 0;
}