hdu1241dfs水题

“

`Oil Deposits

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

这道题是自己写的dfs第一道水题，大致题意就是问你连续的@有多少个，我的思路是，以每个点为起点，只要符合条件就往下搜。
- 这个题没什么坑，就是我忽略了一点，它的搜索方向可以是8个，我的思维还局限在走迷宫最短路径那道例题上，导致自己一直搜索为4个方向，而且只要走过的就标记为‘*’，不用再退回了，这不是找最优路径。

上ac代码

#include
#include
#define N 100

char str[N+10][N+10];
int book[N+10][N+10];
int n,m,count=0;

void dfs(int x,int y)
{
    int i;
    int k[8][2]={{0,1},{0,-1},{1,0},{-1,0},{1,1},{-1,1},{1,-1},{-1,-1}};
    int tx,ty;


    for( i = 0; i < 8; i ++)//搜索8个方向，不止四个方向
    {
        tx = k[i][0] + x;
        ty = k[i][1] + y;
        if(tx < 0||tx > n-1||ty < 0||ty > m-1 )
            continue;
        if( str[tx][ty] == '@'&&book[tx][ty] != '*')
        {
            book[tx][ty] = '*';
            dfs(tx,ty);

        }

    }
    return ;
}

int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m),m!=0)
    {
        getchar();
        count = 0;
        memset(book,0,sizeof(book));
        memset(str,0,sizeof(str));
        for( i =0; i < n; i ++)
        {
            scanf("%s",str[i]);
        }


        for(i = 0; i < n; i ++)
            for( j = 0; j < m; j ++)
            {
                if(str[i][j] == '@'&&book[i][j] != '*')
                {   
                    count++;
                    book[i][j] = '*';
                    dfs(i,j);
                }
            }

        printf("%d\n",count);

    }
    return 0;
}

“`

• 这道题自己想了两天吧，一直没思路（好笨），昨天终于能写了，却满目的bug，今天早上终于无奈了，自己原则是超过3天没想法就看下解题报告，结果发现自己方向一直考虑错了（摊手）。