杭電oj 1241

最近在準備藍橋杯 聽了acm組的建議臨時報佛腳準備學下dfs與bfs 然後看别人的部落格裡有提到一道入門的深搜題目 題目如下：

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

1

2

2

大概意思就是找有幾個油田 @周圍八個格子相連的@隻算同一個油田。

感覺跟以前做的一道連連看的題目有點相似

這題就是通過一個遞歸完成 先周遊每次找到@的地方就對他進行搜尋操作 搜尋旁邊是否有相連的@并且把走過的@變成*

代碼如下：

#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdio>

using namespace std;

char d[120][120];

int n,m;

void dfs(int x,int y)
{
	d[x][y]='*';
	int dx,dy;
	for(int i=-1;i<=1;i++){
		for(int k=-1;k<=1;k++){
			dx=x+i;
			dy=y+k;
			if(dx>=0&&dx<n&&dy>=0&&dy<m&&d[dx][dy]=='@'){
				dfs(dx,dy);
			}
		}
	}
}



int main()
{
	int ans=0,i,k;
	while(scanf("%d%d",&n,&m)!=EOF){
		ans=0;
		getchar();
		if(n==0&&m==0) break;
		for(i=0;i<n;i++){
			for(k=0;k<m;k++){
				scanf("%c",&d[i][k]);
			}
			getchar();
		}
		for(i=0;i<n;i++){
			for(k=0;k<m;k++){
				if(d[i][k]=='@'){
				dfs(i,k);
				ans++;
				}
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}