poj 1111 Image Perimeters

Image Perimeters
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 8573 Accepted: 5130

Description
Technicians in a pathology lab analyze digitized images of slides. Objects on a slide are selected for analysis by a mouse click on the object. The perimeter of the boundary of an object is one useful measure. Your task is to determine this perimeter for selected objects.

The digitized slides will be represented by a rectangular grid of periods, ‘.’, indicating empty space, and the capital letter ‘X’, indicating part of an object. Simple examples are

XX Grid 1 .XXX Grid 2

XX .XXX

              .XXX 

              ...X 

              ..X. 

              X... 

An X in a grid square indicates that the entire grid square, including its boundaries, lies in some object. The X in the center of the grid below is adjacent to the X in any of the 8 positions around it. The grid squares for any two adjacent X’s overlap on an edge or corner, so they are connected.

XXX

XXX Central X and adjacent X’s

XXX

An object consists of the grid squares of all X’s that can be linked to one another through a sequence of adjacent X’s. In Grid 1, the whole grid is filled by one object. In Grid 2 there are two objects. One object contains only the lower left grid square. The remaining X’s belong to the other object.

The technician will always click on an X, selecting the object containing that X. The coordinates of the click are recorded. Rows and columns are numbered starting from 1 in the upper left hand corner. The technician could select the object in Grid 1 by clicking on row 2 and column 2. The larger object in Grid 2 could be selected by clicking on row 2, column 3. The click could not be on row 4, column 3.

One useful statistic is the perimeter of the object. Assume each X corresponds to a square one unit on each side. Hence the object in Grid 1 has perimeter 8 (2 on each of four sides). The perimeter for the larger object in Grid 2 is illustrated in the figure at the left. The length is 18.

Objects will not contain any totally enclosed holes, so the leftmost grid patterns shown below could NOT appear. The variations on the right could appear:

Impossible Possible

XXXX XXXX XXXX XXXX

X..X XXXX X… X…

XX.X XXXX XX.X XX.X

XXXX XXXX XXXX XX.X

….. ….. ….. …..

..X.. ..X.. ..X.. ..X..

.X.X. .XXX. .X… …..

..X.. ..X.. ..X.. ..X..

….. ….. ….. …..

Input
The input will contain one or more grids. Each grid is preceded by a line containing the number of rows and columns in the grid and the row and column of the mouse click. All numbers are in the range 1-20. The rows of the grid follow, starting on the next line, consisting of ‘.’ and ‘X’ characters.

The end of the input is indicated by a line containing four zeros. The numbers on any one line are separated by blanks. The grid rows contain no blanks.

Output
For each grid in the input, the output contains a single line with the perimeter of the specified object.

Sample Input

2 2 2 2
XX
XX
6 4 2 3
.XXX
.XXX
.XXX
…X
..X.
X…
5 6 1 3
.XXXX.
X….X
..XX.X
.X…X
..XXX.
7 7 2 6
XXXXXXX
XX…XX
X..X..X
X..X…
X..X..X
X…..X
XXXXXXX
7 7 4 4
XXXXXXX
XX…XX
X..X..X
X..X…
X..X..X
X…..X
XXXXXXX
0 0 0 0

Sample Output

8
18
40
48
8

Source
Mid-Central USA 2001

题目链接：http://poj.org/problem?id=1111
题目大意：输入一个n*m的字符矩阵，给出起始点坐标，求出与起始点“相连”的’#’组成的图形的周长，其中，“相连”是指东，西，南，北，东北，东南，西北，西南八个方向。

解题思路：简单的深度搜索，枚举八个方向，初始边长为0，如果是边缘或者是字符’.’，边长就加1，如果是’#’，进入下一层，注意标记visit表示访问过，因为不是多条路径的“探索”问题，因此visit不用再恢复0。

代码如下：

#include <cstdio>
#include <cstring>
int n,m,ans;
char s[30][30];
int vis[30][30];
int dx[8]={-1,-1,0,1,1,1,0,-1};   //枚举八个方向
int dy[8]={0,1,1,1,0,-1,-1,-1};
void dfs(int x,int y)
{
    for(int i=0;i<8;i++)
    {
        int nx=x+dx[i];
        int ny=y+dy[i];
        if(i%2==0)
        {
            if(nx<0 || ny<0 || nx>=n || ny>=m || s[nx][ny]=='.')
            {
                ans++;
                continue;
            }
            if(vis[nx][ny])
                continue;
            vis[nx][ny]=1;
            dfs(nx,ny);
        }
        else
        {
            if(nx<0 || ny<0 || nx>=n || ny>=m || s[nx][ny]=='.')
                continue;
            if(vis[nx][ny])
                continue;
            vis[nx][ny]=1;
            dfs(nx,ny);
        }
    }

}
int main()
{
    int sx,sy;
    while(scanf("%d%d%d%d",&n,&m,&sx,&sy)!=EOF && (n||m||sx||sy))
    {
        ans=0;
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++)
            scanf("%s",s[i]);
        vis[sx-1][sy-1]=1;
        dfs(sx-1,sy-1);
        printf("%d\n",ans );
    }
    return 0;
}