【POJ 1979】Red and Black

Red and Black

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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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

代码：
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

char map[30][30];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
bool vis[30][30];
int ans;
int n,m,xs,ys;

void dfs( int x,int y )
{
int i;
for( i=0; i<4; i++ )
{
int nx = x+dx[i];
int ny = y+dy[i];
if( map[nx][ny] == '.' && vis[nx][ny] == false && nx>=0 && ny>=0 && nx<m && ny<n )
{
vis[nx][ny]=true;
ans++;
dfs(nx,ny);
}
}
}
int main()
{
while( ~scanf("%d%d",&n,&m) , n||m )
{
int i,j;
memset(vis,0,sizeof(vis));
for( int i=0; i<m; i++ )
{
getchar();
for( int j=0; j<n; j++ )
{
scanf("%c",&map[i][j]);
if( map[i][j] == '@')
{
xs = i;
ys = j;
vis[i][j] = true;
}
}
}
ans=0;
dfs(xs,ys);
printf("%d\n",ans+1);
}
}

“`

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第一道DFS 纪念一下，最多可以走多少步，回溯，每次走到走不下去了再回来，难的是模板，用一个dx[4]={1，-1,0,0}和dy[4]={0,0，-1,1}来模拟前后左右移动的情况。

由于格式问题，代码发在上面

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