题目描述 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 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 are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.
大致题意 给你一个由*和@组成的图，问你连通区域块的数量
思路 dfs即可
代码如下
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <fstream> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <queue> #include <stack> #include <vector> #include <list> #include<sstream> #include<ctime> using namespace std; char map[101][101]; int dirx[8]={0,0,1,-1,1,-1,1,-1}; int diry[8]={1,-1,0,0,1,-1,-1,1}; int n, m, sum; void dfs(int x, int y) { int x1,y1; map[x][y]='0'; //标记走过 for(int i=0;i<8;i++) { x1=x+dirx[i]; y1=y+diry[i]; if(map[x1][y1]=='@'&&x1>=0&&x1<m&&y1>=0&&y1<n) //满足条件的话继续dfs dfs(x1,y1); } } int main() { int i,j; while(cin>>m>>n) { if(m==0||n==0) break; sum=0; for(i=0;i<m;i++) for(j=0;j<n;j++) cin>>map[i][j]; for(i=0;i<m;i++) { for(j=0;j<n;j++) { if(map[i][j]=='@') { dfs(i,j); sum++; //当这个点dfs完后，数量加一 } } } cout<<sum<<endl; } return 0; }