ACM集训第二天

两天的集训下来在慢慢的摸索过程中逐渐有了一点领会。

先是讲的dfs深度优先搜索法，然后是bfs广度优先搜索法，但是在代码书写的过程却总是出现一些问题。

DFS的主要思想是利用递归，每次递归都会带入下一个节点数据。而BFS则是使用队列或是优先队列进行广度优先的搜索方式。

以下列出一道杭电的题目。

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13991    Accepted Submission(s): 8044

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

 

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这道题其实是一道简化了的紫书上的例题，这道题就是用DFS的一道典型的题目。而在处理dfs的时候一定要注意它的return而且一定先标记，就像用BFS一样，一定是先做好修改在进行push。

下面附上十分拙劣的源码：

#include<cstdio>
#define maxn1 105
#define maxn2 105

using namespace std;

char map[maxn1][maxn1];
int oil[maxn1][maxn1];
int mov[8][2] = {-1,0,1,0,0,-1,0,1,-1,-1,-1,1,1,-1,1,1};
int m,n;
void dfs(int x,int y){
			for(int i = 0;i < 8;i++){
				if(x+mov[i][0] >= 0&& x+mov[i][0] < n&&y + mov[i][1]>=0 && y + mov[i][1] < m){
					if(oil[y+mov[i][1]][x+mov[i][0]]){
						oil[y+mov[i][1]][x+mov[i][0]] = 0;
						dfs(x+mov[i][0],y+mov[i][1]);
					}else
						continue;
				}else
					continue;
			}
		}
int main(){
	while(scanf("%d %d",&m,&n) && m && n){
		for(int i = 0;i < m;i++){
			for(int j = 0;j < n;j++){
				scanf(" %c",&map[i][j]);
				if(map[i][j] == '@')
					oil[i][j] = 1;
				else if(map[i][j] == '*')
					oil[i][j] = 0;
			}
		}
		int cnt = 0;
		for(int i = 0;i < m;i++){
			for(int j = 0;j < n;j++){
				if(oil[i][j]){
					cnt += 1;
					dfs(j,i);
				}
			}
		}
		printf("%d
",cnt);
	}
	return	0;
}