POJ 1562 Oil Deposits

                                                                                                                          Oil Deposits

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 18167 Accepted: 9638

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 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

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

问你有多少个油田，就是给出的字符@，如果这个@在上，下，左，右，左上，左下，右上，右下都有@，就认为他们是连在一起的，所以第二个样例中只有一个油田，因为看中间的那个@，它的左上左下右上右下都有@。

这道题要用深搜，对整个地图遍历一遍，把经过的@都变为其它符号，下一次就遍历不到了。

#include<iostream>#include<cstdio>using namespace std;char map[110][110];int n,m;void dfs(int x,int y){if(x<0||y<0||x>=n||y>=m||map[x][y]!='@') return;else{map[x][y]='*';dfs(x+1,y);dfs(x-1,y);dfs(x,y+1);dfs(x,y-1);dfs(x+1,y+1);dfs(x+1,y-1);dfs(x-1,y+1);dfs(x-1,y-1);}}int main(){while(~scanf("%d%d",&n,&m)){getchar();if(!n&&!m) break;int sum=0;for(int i=0;i<n;i++){for(int j=0;j<m;j++){scanf("%c",&map[i][j]);}getchar();}for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(map[i][j]=='@'){sum++;dfs(i,j);}}}cout<<sum<<endl;}}