InputThe 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.
OutputFor 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：多组输入，输入为两个0时结束输入，对于输入的油田中，‘*’为没有油气的，‘@’为有油气，而当，两个有油气的格子相邻时，（不论是横是竖还是斜）认为它们在同一块
油田内，输出该用例中有几块油田。
思路：深搜，每到一个有油气的格子，替换掉信息，遍历八个方向即可。代码如下：
#include <iostream> #include <cstring> #include <algorithm> #include <string.h> using namespace std; const int max_n=103; char mp[max_n][max_n]; int m,n,sum; bool judge(int x,int y) { return x>=0&&x<m&&y>=0&&y<n; } void dfs(int x,int y) { mp[x][y]=‘*‘;//替换当前位置 for(int dx=-1;dx<=1;dx++)//遍历周围的八个格子 { for(int dy=-1;dy<=1;dy++) { int nx=x+dx,ny=y+dy; if(judge(nx,ny)&&mp[nx][ny]==‘@‘)dfs(nx,ny);//当有油气时，递归 } } return; } int main() { while(scanf("%d %d",&m,&n)==2) { if(m==0&&n==0)break; sum=0; for(int i=0;i<m;i++) { cin>>mp[i]; } for(int i=0;i<m;i++) { for(int j=0;j<n;j++) { if(mp[i][j]==‘@‘)//当有油气时 { dfs(i,j);//替换它以及他周围的同一油田的格子 sum++;//计数器加1 } } } printf("%d\n",sum); memset(mp,0,sizeof(mp)); } return 0; }
原文：https://www.cnblogs.com/whocarethat/p/11100018.html