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
   
   
   
   

 

Source

Mid-Central USA 1997

题意：计算的@大范围有几个（方向有8个方向），比如拿最后一个例子来讲：第一个大范围的@有8个，第二个大范围的@有5个，如此，

解题思路：dfs搜索，记录大范围有几个.

#include<stdio.h>
#include<string.h>
int n,m; int ans; int tx,ty; char map[101][101]; int dir[8][2]={0,1,0,-1,1,0,-1,0,1,1,-1,-1,-1,1,1,-1};//定义8个方向 void dfs(int x,int y) { int i; for(i=0;i<8;i++) {
        tx=x+dir[i][0];
        ty=y+dir[i][1]; if(tx>=1&&tx<=n&&ty>=1&&ty<=m&&map[tx][ty]=='@')//判断边界问题以及位置的信息 {
            map[tx][ty]='*'; //找到@之后把当前位置的@转化为*，不然会一直搜索下去，重复计算。
            dfs(tx,ty); } } } int main() { while(scanf("%d%d",&n,&m)!=EOF) {
        ans=0; if(n==0&&m==0) break; int i,j;
        getchar(); for(i=1;i<=n;i++) { for(j=1;j<=m;j++) {
                scanf("%c",&map[i][j]); }
            getchar(); } for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if(map[i][j]=='@') {    
                    map[i][j]='*';// 找到@之后开始搜索。每次搜索之后总会把在一个范围内所有的@（被搜到的@）被标记为*所以不存在重复计算的问题
                    ans++; //ans++ 
                    dfs(i,j); } } }
        printf("%d\n",ans); 输出ans } return 0; }