Poj 1979 Red and Black

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Language:

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 24921		Accepted: 13447

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)
The end of the input is indicated by a line consisting of two zeros.

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

Japan 2004 Domestic

完全是图的遍历，求所能在局限范围内走的最大步数。

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

char cnt[21][21];
int step,Max[5],W,H;
int vis[21][21];

int dic[4][2]= {{0,1},{1,0},{-1,0},{0,-1}};
void  dfs (int x,int y)
{
    for(int i=0; i<4; i++)
    {
        int tx=x+dic[i][0];
        int ty=y+dic[i][1];
        if(tx>=0&&ty>=0&&tx<H&&ty<W&&!vis[tx][ty]&&cnt[tx][ty]!='#')
        {
            step++;
            vis[tx][ty]=1;
            dfs(tx,ty);

        }
    }
    return ;
}

int main()
{
    int si,sj;
    while(cin>>W>>H,W+H)
    {
        memset(vis,0,sizeof(vis));
        step=1;
        for(int i=0; i<H; i++)
            for(int j=0; j<W; j++)
            {
                cin>>cnt[i][j];
                if(cnt[i][j]=='@')
                {
                    si=i;
                    sj=j;
                }
            }
        vis[si][sj]=1;
        dfs(si,sj);

        cout<<step<<endl;

    }
}

posted @ 2015-05-05 12:37 cbam 阅读(100) 评论(0) 收藏举报

cbam

Poj 1979 Red and Black

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