Oil Deposits

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

算法：深搜

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int MAX = 1e2 + 10;typedef long long LL;char s[MAX][MAX];int n,m,vis[MAX][MAX],ans;int fx[8] = {0,0,-1,1,-1,1,-1,1},fy[8] = {-1,1,0,0,-1,-1,1,1}; // 8 个方向 void dfs(int x,int y){    vis[x][y] = 1;    for(int i = 0; i < 8; i++){        int xx = x + fx[i],yy  = fy[i] + y;        if(xx >= 0 && yy >= 0 && xx < n && yy < m && !vis[xx][yy] && s[xx][yy] == '@') // 判断是否出界， 是否已经搜索过， 是否为油田             dfs(xx,yy);    }}int main(){    while(~scanf("%d %d",&n,&m),m){        memset(vis,0,sizeof(vis));        for(int i = 0; i < n; i++)            scanf("%s",s[i]);        ans = 0;        for(int i = 0; i < n; i++)            for(int j = 0; j < m; j++)                if(!vis[i][j] && s[i][j] == '@')                    ans++,dfs(i,j); // 第一次搜到 + 1         printf("%d\n",ans);    }    return 0;}

广搜：

#include<cstdio>#include<queue>#include<cstring>#include<algorithm>using namespace std;const int MAX = 1e2 + 10;typedef long long LL;char s[MAX][MAX];int n,m,vis[MAX][MAX],ans;int fx[8] = {0,0,-1,1,-1,1,-1,1},fy[8] = {-1,1,0,0,-1,-1,1,1};struct node{    int x,y;};void bfs(int x,int y){    vis[x][y] = 1;    queue <node> q;    node o;    o.x = x,o.y = y;    q.push(o);    while(!q.empty()){        o = q.front();        q.pop();        for(int i = 0; i < 8; i++){ // 8 个方向              int xx = o.x + fx[i],yy = o.y + fy[i];            if(xx >= 0 && yy >= 0 && xx < n && yy < m && !vis[xx][yy] && s[xx][yy] == '@'){  // 判断是否出界， 是否已经搜索过， 是否为油田                 node w;                vis[xx][yy] = 1;//把搜索过的全部都标记上                 w.x = xx,w.y = yy;                q.push(w);            }        }    }}int main(){    while(~scanf("%d %d",&n,&m),m){        memset(vis,0,sizeof(vis));        for(int i = 0; i < n; i++)            scanf("%s",s[i]);        ans = 0;        for(int i = 0; i < n; i++)            for(int j = 0; j < m; j++)                if(!vis[i][j] && s[i][j] == '@')                    ans++,bfs(i,j); // // 第一次搜到 + 1          printf("%d\n",ans);    }    return 0;}

算法三：深度。形式是一样的但是这个算法稍微精炼一些

#include <stdio.h>  int x,y,W,H,num;  char map[111][111];  int ans[5000];  int move_x[8]={0,0,-1,1,1,1,-1,-1};  int move_y[8]={1,-1,0,0,1,-1,-1,1};     //移动  void dfs(int x,int y)  {      if (x<0||x>=H||y<0||y>=W)          return;      if (map[x][y]=='*')          return;      map[x][y]='*'; 将搜索到得全部标记为“*”     for (int i=0;i<8;i++)          dfs(x+move_x[i],y+move_y[i]);  }  int main()  {      while (~scanf ("%d %d",&H,&W) && H)      {          for (int i=0;i<H;i++)          {              scanf ("%s",map[i]);          }          num=0;          for (int i=0;i<H;i++)          {              for (int j=0;j<W;j++)              {                  if (map[i][j]=='@')                  {                      dfs(i,j);                      num++;                  }              }          }          printf ("%d\n",num);      }      return 0;  }