【1241】Oil Deposits（HDU）

Oil DepositsTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d& %I64uSubmit StatusDescriptionThe 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 eachplot 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 numerouspockets. Your job is to determine how many different oil deposits are contained in a grid.InputThe 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 thisare 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.Outputare 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

0
1
2
2

BFS和DFS

#include <stdio.h>#include <string.h>#include <stdlib.h>struct node{int x,y;} q[10010],f,t;int vis[110][110];char ma[110][110];int n,m,num=0;int mv[8][2]= {{0,1},{-1,0},{0,-1},{1,0},{-1,-1},{1,-1},{-1,1},{1,1}};//定义八个方向void dfs(int x,int y){int i;struct node f1;vis[x][y]=1;for(i=0; i<8; i++){f1.x=x+mv[i][0];f1.y=y+mv[i][1];if(f1.x>=0&&f1.x<n&&f1.y>=0&&f1.y<m&&!vis[f1.x][f1.y]&&ma[f1.x][f1.y]=='@'){dfs(f1.x,f1.y);}}}void bfs(int x ,int y){f.x=x;f.y=y;int jin=0,chu=0;int i;q[jin++]=f;while(chu<jin){t=q[chu++];if(ma[t.x][t.y]=='@'&&!vis[t.x][t.y]){dfs(t.x,t.y);//每调用一次，就说明有一个或一群@num++;}for(i=0; i<8; i++){f.x=t.x+mv[i][0];f.y=t.y+mv[i][1];if(f.x>=0&&f.x<n&&f.y>=0&&f.y<m&&!vis[f.x][f.y]){if(ma[f.x][f.y]!='@')vis[f.x][f.y]=1;q[jin++]=f;}}}}int main(){int i,j;while(~scanf("%d %d",&n,&m)&&n!=0&&m!=0){getchar();memset(vis,0,sizeof(vis));memset(ma,0,sizeof(ma));for(i=0; i<n; i++)scanf("%s",ma[i]);for(i=0; i<n; i++){for(j=0; j<m; j++){if(ma[i][j]=='@'&&!vis[i][j])bfs(i,j);}}printf("%d\n",num);num=0;}return 0;}