uva 572 Oil Deposits（DFS遍历图）
| Oil Deposits |
The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil.
A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.
Input
The input file contains one or more grids. Each grid begins with a line containing m and n , the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise and . Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either ` * ', representing the absence of oil, or ` @ ', representing an oil pocket.
Output
For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.
Sample Input
1 1 * 3 5 *@*@* **@** *@*@* 1 8 @@****@* 5 5 ****@ *@@*@ *@**@ @@@*@ @@**@ 0 0
Sample Output
0 1 2 2题目大意：‘@’代表油田，判断给出途中有多少块油田（注意相连的是同一块，相连是八个方向）
解题思路：遍历，碰到‘@’就讲该位置和跟它相邻的‘@’全变成' * ',相当于采集掉了（防止重复计算）。
#include<stdio.h>
#include<string.h>
#define N 105
char map[N][N];
int n, m, cnt;
const int dir[8][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
void DFS(int a, int b){
map[a][b] = '*';
for (int i = 0; i < 8; i++){
if (a + dir[i][0] < 0 || a + dir[i][0] >= n) continue;
if (b + dir[i][1] < 0 || b + dir[i][1] >= m) continue;
if (map[a + dir[i][0]][b + dir[i][1]] == '*') continue;
DFS(a + dir[i][0], b + dir[i][1]);
}
}
int main(){
while (scanf("%d%d%*c", &n, &m), n && m){
// Init.
memset(map, 0, sizeof(map));
cnt = 0;
// Read.
for (int i = 0; i < n; i++)
gets(map[i]);
// Handle.
for (int i = 0; i < n; i ++){
for (int j = 0; j < m; j++){
if (map[i][j] == '@'){
DFS(i, j);
cnt++;
}
}
}
printf("%d\n", cnt);
}
return 0;}