zoukankan      html  css  js  c++  java
  • HDU 1241 Oil Deposits dfs && bfs

    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
     
       下面有代码自己看。
       dfs
     
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int n,m,t,p;
    char map[105][105];
    int yi[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
    void dfs(int x,int y)
    {
        int i,j,rx,ry;
        map[x][y]='*';
        for (i=0;i<8;i++)
        {
             rx=x+yi[i][0];
             ry=y+yi[i][1];
            if (rx>=1 && rx<=n && ry>=1 && ry<=m && map[rx][ry]=='@')
            {
                map[rx][ry]='*';
                dfs(rx,ry);
            }
        }
        return ;
    }
    int main()
    {
        int i,j,ans;
        while (~scanf("%d%d",&n,&m))
        {
            if (n==0&&m==0) break;
            ans=0;
            getchar();
            for (i=1;i<=n;i++)
            for (j=1;j<=m;j++) scanf(" %c",&map[i][j]);
            for (i=1;i<=n;i++)
            for (j=1;j<=m;j++)
            if (map[i][j]=='@')
            {
                dfs(i,j);
                ans++;
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
    

      bfs

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    int n,m;
    char map[105][105];
    int yi[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
    struct f
    {
        int x,y;
    };f s,r;
    void bfs(int x,int y)
    {
        int i,j;
        queue<f>q;
        s.x=x;
        s.y=y;
        map[x][y]='#';
        q.push(s);
        while (!q.empty())
        {
            s=q.front();
            q.pop();
            for (i=0;i<8;i++)
            {
                r.x=s.x+yi[i][0];
                r.y=s.y+yi[i][1];
                if (r.x>=0 && r.x<n && r.y>=0 && r.y<m && map[r.x][r.y]=='@')
                {
                    map[r.x][r.y]='#';
                    q.push(r);
                }
            }
        }
    }
    int main()
    {
        int i,j,ans;;
        while (~scanf("%d%d",&n,&m))
        {
            if (n==0&&m==0) break;
            ans=0;
            for (i=0;i<n;i++)
            for (j=0;j<m;j++) scanf(" %c",&map[i][j]);
            for (i=0;i<n;i++)
            for (j=0;j<m;j++)
            {
                if (map[i][j]=='@')
                {
                    ans++;
                    bfs(i,j);
                }
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
    
  • 相关阅读:
    apache 泛域名配置
    使用Innosetup制作安装包的一些技巧
    以前编写的inno setup脚本,涵盖了自定义安装界面,调用dll等等应用 (转)
    一个比较完整的Inno Setup 安装脚本(转)
    C++(MFC)中WebBrowser去除3D边框的方法(实现IDocHostUIHandler接口)控制 WebBrowser 控件的外观和行为
    computer repair services in Hangzhou
    INNO SETUP 5.5.0以上版本中文语言包
    洛谷P1115 最大子段和【dp】
    洛谷P1996 约瑟夫问题【队列】
    数据结构实验病毒感染检测问题(C++)
  • 原文地址:https://www.cnblogs.com/pblr/p/4696385.html
Copyright © 2011-2022 走看看