题目：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
题意：给定的一个矩阵中，指定一个初始位置，这个初始位置必须是X,这个X周围的8个X都会被感染，被感染的X会继续感染
它周围的8个X,直到无法感染为止。最后要求输出被感染X所构成的连通图形的周长。
输入矩阵行数，列数，初始位置行坐标(从1开始),初始位置列坐标(从1开始)
输入矩阵图形
输出周长
输入 0 0 0 0 时结束程序
示例：
一个6行4列的矩阵，初始位置坐标为2 3，连通图形周长如图所示，周长为18
方法：可以先在初始矩阵周围设置一圈点，DFS遍历，所求图形的周长就是连通的X周围 (上下左右四个方向) 的点的个数
代码
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char M[100][100];
int dir[8][2]={{1,0},{0,1},{0,-1},{-1,0},{1,-1},{-1,-1},{-1,1},{1,1}};
int DFS(int m,int n)
{
int sum=0;
if(M[m][n]=='V'||M[m][n]=='.')return 0;//已经访问过或者是.的情况下，加0
M[m][n]='V';
for(int i=0;i<4;i++)
{
if(M[m+dir[i][0]][n+dir[i][1]]=='.')//上下左右四个方向
sum++;
}
for(int i=0;i<8;i++)
{
sum+=DFS(m+dir[i][0],n+dir[i][1]);//八个方向
}
return sum;
}
int main()
{
int h,l,x,y;
while(cin>>h>>l>>x>>y)
{
int sum;
if(h==0&&l==0)
{
return 0;
}
memset(M,'.',sizeof(M));
for(int i=1;i<=h;i++)
for(int j=1;j<=l;j++)
{
cin>>M[i][j];
}
sum=DFS(x,y);
cout<<sum<<endl;
}
return 0;
}