nta vco c++ logic *** ids containe 文件 nes

HDU_1241 Oil Deposits

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 題目大意：

地質urvcomp地質勘測公司負責探測地下油層。GeoSurvComp每次處理一個大的矩形區域，並創建一個網格，將土地劃分為許多方形地塊。然後用傳感設備分別分析每個地塊，以確定該地塊是否含有石油。一塊含有石油的土地叫做口袋。如果兩個儲層相鄰，則它們屬於同一油層的一部分。石油儲量可以相當大，可能包含許多口袋。你的工作是確定有多少不同的石油蘊藏在一個網格。

輸入

輸入文件包含一個或多個網格。每個網格都以包含m和n的一行開始，m和n是網格中行和列的數量，由一個空格分隔。如果m = 0，則表示輸入結束;否則1 <= m <= 100, 1 <= n <= 100。下面是m行，每行有n個字符(不包括行尾字符)。每個角色對應一個情節，要麽是“*”，代表沒有油，要麽是“@”，代表一個油袋。

輸出

對於每個網格，輸出不同的石油沈積物的數量。如果兩個不同的儲層在水平、垂直或對角方向相鄰，那麽它們就是同一個油層的一部分。一個石油礦床不會包含超過100個口袋。

樣例輸入 1 1 * 3 5 *@*@* **@** *@*@* 1 8 @@****@* 5 5 ****@ *@@*@ *@**@ @@@*@ @@**@ 0 0 樣例輸出 0 1 2 2

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 char martrix[110][110];
 5 int move[8][2]={1,0,-1,0,0,1,0,-1,1,1,-1,-1,1,-1,-1,1};//兩個坐標一組 分為8組
 6 int h,w;
 7 void dfs(int x,int y)//定義dfs函數，主函數找到了@，dfs啟動，尋找主函數找到的@八面存在的@
 8 {
 9     int next_x,next_y,i;
10     martrix[x][y]=‘*‘;//把找到的@變為*，以免重復搜索
11     for(i=0;i<8;i++)
12     {
13         next_x=x+move[i][0];//[0]表示兩個坐標一組的第一個 [i]表示兩個坐標一組的第幾組
14         next_y=y+move[i][1];//[1]表示兩個坐標一組的第二個 [i]表示兩個坐標一組的第幾組
15         if(next_x>=0&&next_x<h&&next_y>=0&&next_y<w)
16         {
17             if(martrix[next_x][next_y]==‘@‘)
18             {
19                 dfs(next_x,next_y);
20             }
21         }
22     }
23 }
24 
25 int main()//主函數開始，尋找第一個@
26 {
27     freopen("input.txt","r",stdin);
28     int i,j,sum;
29     while(scanf("%d%d",&h,&w)&&(h!=0||w!=0))
30     {
31         for(i=0;i<h;i++)
32         scanf("%s",martrix[i]);
33         sum=0;
34         for(i=0;i<h;i++)
35         {
36             for(j=0;j<w;j++)
37             {
38                 if(martrix[i][j]==‘@‘)
39                 {
40                     dfs(i,j);//轉移到dfs函數，由dfs函數開始搜索主函數找到@的八面存在的@
41                     sum++;
42                 }
43             }
44         }
45         printf("%d\n",sum);
46     }
47     fclose(stdin);
48     return 0;
49 }

（經典DFS）HDU_1241 Oil Deposits