HDOJ 1241 Oil Deposits——邻接矩阵BFS和DFS

The Mine  

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  and . 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

 BFS：

#include <stdio.h>#include <stdlib.h>#include <string.h>char map[110][110];//图的存储int vis[110][110];//标记是否被访问过int num,n,m;struct linshi{    int x;    int y;} ls[604];//用来存储找的符合题意的点的坐标int bhx[]= {-1,-1,0,1,1,1,0,-1};//八个方向x的变化int bhy[]= {0,1,1,1,0,-1,-1,-1};//八个方向y的变化void bfs(int i,int j){    struct linshi t,f;    int a=0,b=0,k=0;    t.x=i;    t.y=j;    ls[b++]=t;    vis[t.x][t.y]=1;    while(a<b)    {        t=ls[a++];//每次都访问一个符合题意的点        for(k=0; k<8; k++)//寻找这个店的八个方向是否符合题意        {            f.x=t.x+bhx[k];            f.y=t.y+bhy[k];            if(f.x>=0&&f.x<n&&f.y>=0&&f.y<m&&(!vis[f.x][f.y])&&map[f.x][f.y]=='@')//判断            {                ls[b++]=f;//符合题意的被存进结构体里                vis[f.x][f.y]=1;//标记这个点已经被访问过了            }        }    }    num++;}int main(){    int i,j;    while(scanf("%d%d",&n,&m),n&&m)    {            memset(map,'\0',sizeof(map));            memset(vis,0,sizeof(vis));            memset(ls,0,sizeof(ls));            num=0;            for(i=0; i<n; i++)            {                scanf("%*c%s",map[i]);            }            for(i=0; i<n; i++)            {                for(j=0; j<m; j++)                {                    if(map[i][j]=='@'&&(!vis[i][j]))                    {                        bfs(i,j);                    }                }            }            printf("%d\n",num);        }    return 0;}

DFS：

这个和上边差不多

#include <stdio.h>#include <stdlib.h>#include <string.h>char map[110][110];int vis[100][110];int num,n,m;int bhx[]= {-1,-1,0,1,1,1,0,-1};int bhy[]= {0,1,1,1,0,-1,-1,-1};void dfs(int i,int j){    int k,a,b;    vis[i][j]=1;    for(k=0;k<8;k++)    {        a=i+bhx[k];        b=j+bhy[k];        if(a>=0&&a<n&&b>=0&&b<m&&(!vis[a][b])&&map[a][b]=='@')        {            dfs(a,b);//每次发现合格的点，就以这个点为起点继续访问        }    }}int main(){    int i,j;    while(scanf("%d%d",&n,&m),n&&m)    {        memset(map,'\0',sizeof(map));        memset(vis,0,sizeof(vis));        num=0;        for(i=0;i<n;i++)        {            scanf("%*c%s",map[i]);        }        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                if(map[i][j]=='@'&&(!vis[i][j]))                {                    dfs(i,j);                    num++;                }            }        }        printf("%d\n",num);    }    return 0;}