Image Perimeters
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 9183 | Accepted: 5426 |
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 Outpu8 18 40 48 8
题意：这道题可以看作一个感染的样例，给定一个矩阵，指定一个母体X的位置，处在这个X周围8个位置内的X都会被感染，被感染的X仍可以感染它周围8个位置，最后直到无法感染位置。要求的就是最后得到的被感染的图形周长。
分析图形可以知道，.相当于被感染X的隔离带，需要求的就是隔离带的长度，被感染的X周围有1个.，周长就会+1，最终的周长就是被感染的X周围.的和。
思路：将输入的矩阵周围用.包一圈，然后用DFS进行遍历，在遍历的同时统计被感染X周围的.数量，遍历结束得到周长
下面给出AC代码：
1 #include<iostream>
2 #include<cstring>
3 using namespace std;
4 char M[30][30];
5 int dir[8][2] = { { 1,0 },{ 0,1 },{ 0,-1 },{ -1,0 },{ 1,-1 },{ -1,-1 },{ -1,1 },{ 1,1 } };//→↑←↓
6 int DFS(int m, int n)
7 {
8 int sum = 0;
9 if (M[m][n] == ‘0‘ || M[m][n] == ‘.‘) return 0;
10 M[m][n] = ‘0‘;
11 for (int i = 0;i < 4;i++)
12 {
13 if (M[m + dir[i][0]][n + dir[i][1]] == ‘.‘)
14 sum++;
15 }
16 for (int i = 0;i < 8;i++)
17 {
18 sum = sum + DFS(m + dir[i][0], n + dir[i][1]);
19 }
20 return sum;
21 }
22 int main()
23 {
24 int l, w, x, y;
25 while (cin >> l >> w >> x >> y)
26 {
27 if (l == 0 && w == 0)
28 return 0;
29 memset(M, ‘.‘, sizeof(M));
30 for (int i = 1;i <= l;i++)
31 for (int j = 1;j <= w;j++)
32 cin >> M[i][j];
33 int sum;
34 sum = DFS(x, y);
35 cout << sum << endl;
36 }
37 return 0;
38 }
这是我第一篇博客，为了庆祝我单身快乐，1111送给自己。
希望各位大大能够对我多指导指导，感谢~！感谢~~！感谢~~~！