Hdu 1241

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17290    Accepted Submission(s): 9961

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

这道题是典型的深度搜索题，可以根据状态的转换来判断是否可以继续搜索。在搜索的过程中把已经搜索过的位置变成不能被搜索的状态，在主函数中记录能进入搜索的次数及为目标块数。（可以全部进行搜索，也可以只对输入的“油田”位置进行搜索）

以下是我的代码:

#include<iostream>#include<cstdio>using namespace std;const int Max=10010;int dx[8]={-1,-1,-1,0,0,1,1,1};int dy[8]={-1,0,1,-1,1,-1,0,1};char d[Max][Max];int M[Max],N[Max];int num,num2,num1,Num;void dfs(int x,int y){    int xx,yy;    d[x][y]='*';    for(int i=0;i<8;i++){        xx=x+dx[i];        yy=y+dy[i];        //位置的转移        if(xx<0||yy<0||xx>=num||yy>=num2) continue;//判断越界        else if(d[xx][yy]=='@') dfs(xx,yy);    }}int main(){    while(scanf("%d%d",&num,&num2)!=EOF){            if(num==0) break;            Num=num1=0;            for(int i=0;i<num;i++){                    scanf("%s",d[i]);                    for(int j=0;j<num2;j++){                    if(d[i][j]=='@')                    {                        M[num1]=i;                        N[num1]=j;                        num1++;                    }//保存输入的“油田位置”                }            }            for(int i=0;i<num1;i++){                if(d[M[i]][N[i]]=='@'){                dfs(M[i],N[i]);//能进入搜索的必然是油田块                Num++;                }            }            printf("%d\n",Num);    }    return 0;}