HDU 1241 Oil Deposits 且行且珍惜--

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Oil Deposits

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1241

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

这是道老题了，做过好多遍了在刚进入acm的时候学长们就出这种题来训练我们，那时候一脸懵逼的样子，现在回想起来现在写绝对没有以前写的时候那么困难了。但是写的时候还是会出一些小错误，前几天写的时候用的那个深搜直接写的，后来感觉写错了这是有八个方向和状态的题目；后来过了几天又重新写了一遍这道题的核心思路就是把相连的油田都换成*然后看看有几个不相连的。记着刚开始写的时候还考虑回溯啊什么的，现在写起来很得心应手。呵呵了。

题意：@表示油田，问有几个成片的油田，也就是油田群有几个。可以上下左右斜着连一共八个方向；

AC代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int to[8][2]{{1,0},{1,1},{1,-1},{-1,0},{-1,1},{-1,-1},{0,1},{0,-1}};//八个方向
int n, m;
char mapp[105][105];//记录符号的
void dfs(int x, int y)
{
    int i, j;
    int X, Y;
    mapp[x][y]='*';//搜到油田就变为*，然后接着搜
    for(i=0;i<8;i++)
    {
        X=to[i][0]+x;
        Y=to[i][1]+y;
        if( X>=0 && Y>=0 && X<n && Y<m && mapp[X][Y]=='@')
            dfs(X,Y);
    }
}
int main()
{
    int i, j, sum;
    while(scanf("%d%d",&n,&m),n+m)
    {
        getchar();
        for(i=0;i<n;i++)///这里我很纠结的 在编译器上输入数据时得出的结果又误差，但是提交的时候对了，
           gets(mapp[i]);
        ///scanf("%s",&mapp[i]);所以用这个试了试
        sum=0;///记录油田群的个数的
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(mapp[i][j]=='@')
                {
                    dfs(i, j);
                    sum++;
                }
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}