HDU 1241 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 

Sample Output

0122

题目大意：

输入两个整数n，m，接下来输入一个n*m的字符矩阵。当n与m都等于0时结束输入。在矩阵中有“*”，代表没有油，或@，代表油袋。当@的八个方向有@时，这两个油袋表示相连是一个。

c++

#include <stdio.h>#include<algorithm>#include<iostream>using namespace std;char mp[101][101];int a,b,s;void dg(int i, int j){    if(mp[i][j]!='@'||i<0||j<0||i>=a||j>=b)    //当找不到油袋或出界时返回        return;    else    {        mp[i][j]='*';      //将找到的油袋全部改为没有油的状态，并查找它的八个方向        dg(i-1, j-1);        dg(i-1, j);        dg(i-1, j+1);        dg(i, j-1);        dg(i, j+1);        dg(i+1, j-1);        dg(i+1, j);        dg(i+1, j+1);    }}int main(){    int i, j;    while(~scanf("%d%d",&a,&b))    {        if(a==0||b==0) break;        s=0;        for(i=0;i<a;i++)        {            for(j=0;j<b;j++)            {                cin>>mp[i][j];            }        }        for(i=0;i<a;i++)        {            for(j=0;j<b;j++)            {                if(mp[i][j] == '@')    //当找到油袋时进行深搜，油袋个数加1；                {                    dg(i, j);                    s++;                }            }        }        printf("%d\n",s);    }    return 0;}