Oil Deposits

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

Total Submission(s) : 2   Accepted Submission(s) : 2

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

0 1 2 2  

Source

Mid-Central USA 1997

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1241

题意:统计字符’@’组成多少个八连块.如果两个字符’@’所在的格子相邻(横,竖或者对角线方向),就说它们属于同一个八连块.

AC代码:

刘汝佳<<算法竞赛入门经典>>例6-12.分析:

该题属于图的遍历.有DFS和BFS.由于DFS更容易编写,一般用DFS找联通块:从每个”@”格子出发.递归遍历它周围的”@”格子.每次访问一个格子的时候就给它写上”连通分量编号”(即下面代码中的idx数组),这样就可以在访问之前检查它是否有了编号,从而避免同一格子访问多次:

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 100+5;
char pic[maxn][maxn];
int m,n,idx[maxn][maxn];

void DFS(int r,int c,int id)
{
    if(r<0||r>=m||c<0||c>=n) return ;
    if(idx[r][c]>0||pic[r][c]!='@') return ;
    idx[r][c]=id;
    for(int dr=-1;dr<=1;dr++)
        for(int dc=-1;dc<=1;dc++)
            if(dr!=0||dc!=0)
                DFS(r+dr,c+dc,id);
}
int main()
{
    while(scanf("%d%d",&m,&n)!=EOF&&m&&n)
    {
        for(int i=0;i<m;i++) scanf("%s",pic[i]);
        memset(idx,0,sizeof(idx));
        int cnt=0;
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                if(idx[i][j]==0&&pic[i][j]=='@')
                    DFS(i,j,++cnt);
        printf("%d\n",cnt);
    }
    return 0;
}

上面的代码用一个二重循环来判来找到当前格子的相邻八个格子,也可以用常量数组或者写8条DFS调用,读者可以根据自己的喜好选用.这道题的算法有个好听的名字:种子填充(floodfill).有兴趣的同学还可以看看维基百科中的动画,对DFS和BFS实现的种子填充有一个更加直观的认识.

DFS

#include<cstdio>
#include <cstring>
using namespace std;
#define maxn 105
char a[maxn][maxn];
bool vis[maxn][maxn];
int n,m;
void DFS(int x,int y)
{
    if(x<0||x>=n||y<0||y>=m) return ;//越界
    if(a[x][y]=='*'||vis[x][y]) return;//非油田||来过
    vis[x][y]=true;
    //搜索周围八个方向
    DFS(x+1,y+1);
    DFS(x+1,y);
    DFS(x+1,y-1);
    DFS(x,y-1);
    DFS(x-1,y-1);
    DFS(x-1,y);
    DFS(x-1,y+1);
    DFS(x,y+1);
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n+m==0) break;
        for(int i=0; i<n; i++)
            scanf("%s",a[i]);
        int sum=0;
        memset(vis,false,sizeof(vis));
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                if(!vis[i][j]&&a[i][j]=='@')
                {
                    DFS(i,j);
                    sum++;
                }
        printf("%d\n",sum);
    }
    return 0;
}

BFS:

/*
解题思路：广度优先搜索，图的连通分量。

遍历图——>找到油田——>把油田群找出来——>标记、计数——>继续遍历——>遍历完成

只要相邻既是一起，也就是说，遍历方向有8个，所有相邻的都算。
*/
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m;
char a[100][100];
int sum;
int dir[8][2]={1,0,1,1,1,-1,0,1,0,-1,-1,0,-1,1,-1,-1};      //8个方向
struct node
{
    int x;
    int y;
};
int OK(int x,int y)
{
    if(x>=0&&x<n&&y>=0&&y<m)
        return true;
    return false;
}
void bfs(int x,int y)
{
    int i;
    node now,next;
    queue<node> q;
    now.x=x;
    now.y=y;
    q.push(now);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        for(i=0;i<8;i++)
        {
            next.x=now.x+dir[i][0];
            next.y=now.y+dir[i][1];
            if(OK(next.x,next.y)&&a[next.x][next.y]=='@')     //是油田哦
            {
                a[next.x][next.y]='*';     //油田群遍历，标记
                q.push(next);
            }
        }
    }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)&&m)
    {
        sum=0;
        for(i=0;i<n;i++)
            scanf("%s",a[i]);
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(a[i][j]=='@')      //找到油田
                {
                    sum++;
                    bfs(i,j);
                }
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}