Oil Deposits(bfs)

Oil Deposits

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 5 Accepted Submission(s) : 4

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：给出一副地图，在地图上有一些采油点“@”表示，与采油点“@”直接相连的“@”统称在为一个油田，问

                    给出的地图中共有多少个油田

解题：bfs

#include<cstdio>#include<cstring>#include<iostream>#include<queue>#define M 110char map[M][M];bool mark[M][M];int move[8][2]={{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};int m,n,i,j,k,counter;struct node{    int x,y;}p,q;using namespace std;int main(){    while(scanf("%d%d",&m,&n),m+n)    {        getchar();//调试了半天总是超时...原来是这里、为什么这里要用这个函数...        for(i=0;i<m;i++)          gets(map[i]);        memset(mark,true,sizeof(mark));        counter=0;        for(i=0;i<m;i++)          for(j=0;j<n;j++)            if(map[i][j]=='@'&&mark[i][j])             {                 counter++,mark[i][j]=false;                 queue<node>Q;                 q.x=i,q.y=j;                 Q.push(q);                 while(!Q.empty())                 {                     q=Q.front(),Q.pop();                     for(k=0;k<8;k++)                     {                         p.x=q.x+move[k][0],p.y=q.y+move[k][1];                         if(p.x>=0&&p.x<m&&p.y>=0&&p.y<n&&map[p.x][p.y]=='@'&&mark[p.x][p.y])                          {                            mark[p.x][p.y]=false;                            Q.push(p);                          }                     }                 }             }      printf("%d\n",counter);    }    return 0;}