TOJ1334

                                                               1334: Oil Deposits

时间限制(普通/Java):1000MS/10000MS     内存限制:65536KByte
总提交: 555            测试通过:404

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

题目来源

Mid-Central USA 1997

题解：

题目意思是判断油带的数量，相邻油田属于同一油带，相邻，是指两个小正方形区域上下、左右、左上右下或左下右上同为’@’。

简单深搜。

#include<iostream>
using namespace std;
int x,y,i,j,s;
char arr[101][101];
int g[8]={0,0,1,1,1,-1,-1,-1};
int q[8]={1,-1,0,-1,1,0,1,-1};
int dfs(int a,int b)
{
if(a<0||b<0||a>=x||b>=y)return 0;
if(arr[a][b]=='*')return 0;
arr[a][b]='*';
for(int k=0;k<8;k++)
{
dfs(a+g[k],b+q[k]);
}
}
int main()
{
while(cin>>x>>y)
{
if(x==0&&y==0)break;
s=0;
for(i=0;i<x;i++)
{
for(j=0;j<y;j++)
{
cin>>arr[i][j];
}
}
for(i=0;i<x;i++)
{
for(j=0;j<y;j++)
{
if(arr[i][j]=='@')
{
s++;
dfs(i,j);
}
}
}
printf("%d\n",s);
}
}