hdu 1241 Oil Deposits 解题报告

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9599 Accepted Submission(s): 5632

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
告诉你一个m*n的矩阵，里面有@字符和*字符，其中@代表油井。题目认为一个油井在其8个方位若也含有油井，则归为一个油井。输出油井的个数。
题目做法
经典的DFS
解题代码
#include <iostream>#include <stdio.h>using namespace std;char map[105][105];int n,m,t;int d[8][2]= {0,-1,0,1,1,0,-1,0,-1,-1,1,1,1,-1,-1,1};void dfs(int x,int y){    if(x>=0 && x<m && y>=0 && y<=n)        if(map[x][y]=='@')        {            map[x][y]='*';            for(int i=0; i<8; i++)//  对每一个油井对它的8个方位进行搜索            {                int nx=x+d[i][0];                int ny=y+d[i][1];                dfs(nx,ny);            }        }}int main(){    while(1)    {        scanf("%d%d",&m,&n);        if(m==0&&n==0)            return 0;        t=0;        for(int i=0; i<m; i++)            scanf("%s",map[i]);        for(int i=0; i<m; i++)            for(int j=0; j<n; j++)            {                if(map[i][j]=='@')                {                    t++;                    dfs(i,j);                }            }        printf("%d\n",t);    }    return 0;}