HDU-1241 Oil Deposits

HDU-1241 Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)
 

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

 
题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=1241
 

分析

题意：‘@’表示一个油田，如果‘@’周围也有油田的话，则表示这是一个油田的一部分，就是表示为一个大油田，而这道题就是让我们找出有多少个大油田
 
思路：很明显，这是一道深搜题，从地图的第一个开始搜索，找到一个’@’后，对其周围8个方向进行搜索，如果是油田的话就标记为’*’，直到8个方向都不是’@’，然后使油田数+1就行了，最后输出这个数。
 

代码

#include <iostream>#include <stdio.h>using namespace std;int xy[8][2]={1,0,1,1,0,1,-1,1,-1,0,-1,-1,0,-1,1,-1};int num;int m,n;char dt[100][100];void DFS(int x,int y);int main(){    while (~scanf("%d %d",&m,&n),m!=0&&n!=0)    {        num=0;        for (int i=0;i<m;i++)        {            for (int j=0;j<n;j++)               cin>>dt[i][j];        }        for (int i=0;i<m;i++)        {            for (int j=0;j<n;j++)            {                if (dt[i][j]=='@')                {                    DFS(i,j);                    num++;                }            }        }        cout<<num<<endl;    }    return 0;}void DFS(int x,int y){    int X,Y;    if (dt[x][y]!='@'||x>=m||y>=n||x<0||y<0)        return;    for(int i=0;i<8;i++)    {        dt[x][y]='*';        X=x+xy[i][0];        Y=y+xy[i][1];        DFS(X,Y);    }}