题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1241
Oil Deposits
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13137 Accepted Submission(s):
7611
Problem Description
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 1 <= m <= 100 and 1 <= n <= 100. 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：
这个题目就是要查找油田的个数，比如说第二组数据
3 5
*@*@*
**@** 注明："@"这个与他的几个方向都是@都是可以相连的。所以称为一块油田。输出1。
*@*@*
这种就很容易想到搜索。这里的一个技巧就是起点就从@开始，其次就是找到@就使其变成*；不过找过的还是要标记为1，否则就会重复，这样不连接的还可以继续搜~
东西还是要反复的咀嚼，不然就会很生，忘得差不多0.0
详见代码。
1 #include <iostream>
2 #include <cstdio>
3 #include <queue>
4 #include <cstring>
5
6 using namespace std;
7
8 int dir[8][2]= {0,1,0,-1,1,0,-1,0,1,1,1,-1,-1,1,-1,-1};
9 char map[110][110];
10 int a,b,vis[110][110],k;
11
12 struct node
13 {
14 int x,y;
15 int t;
16 } s,ss;
17
18 queue<node>q,qq;
19 int bfs()
20 {
21 while (!q.empty())
22 {
23 s=q.front();
24 q.pop();
25 //vis[s.x][s.y]=1;
26 for (int i=0; i<8; i++)
27 {
28 int x=s.x+dir[i][0];
29 int y=s.y+dir[i][1];
30 //int t=s.t+1;
31 if (x>=0&&x<a&&y>=0&&y<b)
32 {
33 if (!vis[x][y]&&map[x][y]==‘@‘)
34 {
35 ss.x=x;
36 ss.y=y;
37 map[ss.x][ss.y]=‘*‘;
38 //vis[x][y]=1;
39 q.push(ss);
40 }
41
42 }
43 }
44 }
45 }
46
47 int main ()
48 {
49 while (scanf("%d%d",&a,&b)!=EOF)
50 {
51 memset(vis,0,sizeof(vis));
52 if (a==0&&b==0)
53 break;
54 for (int i=0; i<a; i++)
55 {
56 getchar();
57 for (int j=0; j<b; j++)
58 {
59 scanf("%c",&map[i][j]);
60 }
61 }
62 //int k=0;
63 for (int i=k=0; i<a; i++)
64 {
65 for (int j=0; j<b; j++)
66 {
67 if (map[i][j]==‘@‘)
68 {
69 k++;
70 s.x=i;
71 s.y=j;
72 map[s.x][s.y]=‘*‘;
73 vis[s.x][s.y]=1;
74 q.push(s);
75 bfs();
76 }
77 }
78 }
79 printf ("%d\n",k);
80 }
81 return 0;
82 }