zoukankan      html  css  js  c++  java
  • POJ 3620 Avoid The Lakes (求连接最长的线)(DFS)

    Description

    Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of water. His insurance agency will only repay him, however, an amount depending on the size of the largest "lake" on his farm.

    The farm is represented as a rectangular grid with N (1 ≤ N ≤ 100) rows and M (1 ≤ M ≤ 100) columns. Each cell in the grid is either dry or submerged, and exactly K (1 ≤ K ≤ N × M) of the cells are submerged. As one would expect, a lake has a central cell to which other cells connect by sharing a long edge (not a corner). Any cell that shares a long edge with the central cell or shares a long edge with any connected cell becomes a connected cell and is part of the lake.

    Input

    * Line 1: Three space-separated integers: NM, and K
    * Lines 2..K+1: Line i+1 describes one submerged location with two space separated integers that are its row and column: R and C

    Output

    * Line 1: The number of cells that the largest lake contains. 

    Sample Input

    3 4 5
    3 2
    2 2
    3 1
    2 3
    1 1

    Sample Output

    4
    给出一串坐标,连在一起的算一个,计算每个由多少个坐标组成,输出最大值。
     1 #include<cstdio>
     2 #include<string.h>
     3 int dx[4]={-1,1,0,0};
     4 int dy[4]={0,0,-1,1};
     5 int n,m,map[150][150],x,y,i,max,ans,k,j;
     6 void f(int x,int y)
     7 {
     8     int i,nx,ny;
     9     for(i = 0 ; i < 4 ; i++)
    10     {
    11         nx=x+dx[i];
    12         ny=y+dy[i];
    13         if(nx > 0 && ny > 0 && nx <= n && ny <=m && map[nx][ny] == 1)
    14         {
    15             ans++;
    16             map[nx][ny]=0;
    17             f(nx,ny);
    18         }
    19     }
    20 }
    21 int main()
    22 {
    23     while(scanf("%d %d %d",&n,&m,&k)!=EOF)
    24     {
    25         max=0;
    26         memset(map,0,sizeof(map));
    27         for(i = 0 ; i < k ; i++)
    28         {
    29             scanf("%d %d",&x,&y);
    30             map[x][y]=1;    //1表示没走过 
    31         }
    32         for(i = 1 ; i <=n  ; i++)
    33         {
    34             for(j = 1 ; j <= m ; j++)
    35             {
    36                 
    37                 if(map[i][j] == 1)
    38                 {    
    39                     ans=1;
    40                     map[i][j]=0;
    41                     f(i,j);
    42                     max=max>ans?max:ans;
    43                 }
    44                 
    45             }
    46             
    47         }
    48         printf("%d
    ",max);
    49     }
    50 }
    ——将来的你会感谢现在努力的自己。
  • 相关阅读:
    转:页面Postback后定位滚动条不再难
    c:\windows\microsoft.net\framework\v1.1.4322\Config\machine.config 行: 198
    WebService相关概念和原理(中间层)
    JS 根据DropDownList的Text选中某一项
    javascript事件列表解说
    AJAXUpdateProgress设置CSS元素POSITION的使动画居中 & loading的Info
    ASP.NET2.0 Skin+CSS 测试
    C# 日期格式转换(转)
    编写代码创建DataTable对象
    ToString 格式化数值
  • 原文地址:https://www.cnblogs.com/yexiaozi/p/5715021.html
Copyright © 2011-2022 走看看