POJ 1562 DFS水题练习

最新推荐文章于 2021-07-14 20:05:29 发布

最新推荐文章于 2021-07-14 20:05:29 发布 · 75 阅读

本文介绍了一个用于检测地下油藏的算法。该算法通过创建网格来划分土地并分析每个地块是否含有油藏。若两个油藏相邻，则认为属于同一油藏。文章提供了完整的源代码实现，包括深度优先搜索算法。

Oil Deposits

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6282		Accepted: 3557

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 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.

Sample Input

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5 
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

Sample Output

#include<stdio.h>
#include<stdlib.h>

int m,n;
char grid[102][102];

void DFS(int a,int b)
{
    if(grid[a][b]=='*'||a<0||b<0||a>m-1||b>n-1) return;
    grid[a][b]='*';
	//printf("访问grid[%d][%d]/n",a,b);
        DFS(a,b+1);
        DFS(a,b-1);
        DFS(a-1,b-1);
        DFS(a-1,b+1);
        DFS(a-1,b);
        DFS(a+1,b-1);
        DFS(a+1,b);
        DFS(a+1,b+1);

}
int main()
{
    int i,j;
    int cnt;
    while(scanf("%d %d",&m,&n)!=EOF&&m!=0)
    {
        cnt=0;
        memset(grid,0,sizeof(grid));
        for(i=0;i<m;i++)
            scanf("%s",grid[i]);
         for(i=0;i<m;i++)
            for(j=0;j<n;j++)
            {
                if(grid[i][j]=='@')
                {
                    cnt++;
                    DFS(i,j);
                }
            }
        printf("%d/n",cnt);
    }
    return 0;
}