Oil Deposits

                                                                                   Oil Deposits

时间限制(普通/Java):3000MS/10000MS          运行内存限制:65536KByte
总提交:158            测试通过:116

描述

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.

输入

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.

输出

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.

样例输入

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

样例输出

0
1
2
2

这道题的大概意思是：@代表油田，*代表没有油田，如果一个油田8个方向有油田，则它们属于同一块区域。要求出有多少个区域。

这道题经典的深度搜索题。

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int n,m;char map[110][110];//储存输入数据int vis[8][2]= {{-1,-1},{-1,0},{-1,1},{0,1},{0,-1},{1,1},{1,0},{1,-1}};//分表代表8个方向。void dfs(int x,int y){    int i;    map[x][y]='*';//搜索过了设置成*保证不会经过    for(i=0; i<8; i++)//枚举8种走发    {        //计算下一个点的坐标        int dx=x+vis[i][0];        int dy=y+vis[i][1];        if(dx<0||dx>=n||dy<0||dy>=m)//判断是否越界            continue;        if(map[dx][dy]=='@')        {            dfs(dx,dy);//开始尝试下一个点        }    }}int main(){    int i,j;    while(scanf("%d %d",&n,&m)!=EOF)    {        if(n==0&&m==0)            break;        for(i=0; i<n; i++)        {            scanf("%s",map[i]);        }        int sum=0;        for(i=0; i<n; i++)        {            for(j=0; j<m; j++)            {                if(map[i][j]=='@')                {                    sum++;//数目加一                    dfs(i,j);                }            }        }        printf("%d\n",sum);    }    return 0;}