zoukankan      html  css  js  c++  java
  • Poj3620--Avoid The Lakes(基础Dfs)

    Avoid The Lakes
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 6908   Accepted: 3680

    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 ≤ KN × 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: N, M, 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

    Source

     
    RE: 败给了读题, < 被标记的是湿地 >,
     1 #include <cstdio>
     2 #include <cstring>
     3 #include <iostream>
     4 using namespace std;
     5 int map[1010][1010], ac[4][2] = {1, 0, -1, 0, 0, 1, 0, -1}, ant;
     6 int Dfs(int x, int y,int m, int n)
     7 {
     8     int nx, ny;
     9     map[x][y] = 0;
    10     ant++;
    11     for(int i=0; i<4; i++)
    12     {
    13         nx = x + ac[i][0];
    14         ny = y + ac[i][1];
    15         if(nx>=1 && nx<=m && ny>=1 && ny <= n && map[nx][ny])
    16         {
    17                 Dfs(nx, ny, m, n); 
    18         }
    19     }    
    20 } 
    21 int main()
    22 {
    23     int m, n, t;
    24     while(~scanf("%d %d %d", &m, &n, &t))
    25     {
    26         for(int i = 1; i <= m; i++)
    27             for(int j = 1; j <= n; j++)
    28                 map[i][j] = 0;
    29         int a, b;
    30         for(int i=1; i<=t; i++)
    31         {
    32             scanf("%d %d", &a, &b);
    33             map[a][b] = 1;    
    34         }    
    35         int max = 0;
    36         for(int i=1; i<=m; i++)
    37         {
    38             for(int j=1; j<=n; j++)
    39             {
    40                 if(map[i][j])
    41                 {
    42                     ant = 0;
    43                     Dfs(i, j, m, n);
    44                     if(ant > max)
    45                         max = ant;
    46                 }
    47             }
    48         }
    49         printf("%d
    ", max);
    50     }
    51     return 0;
    52 } 
  • 相关阅读:
    使用Spring的ReloadableResourceBundleMessageSource读取properties配置
    IFrame自适应高度
    Js返回页面顶部
    复制页面内容时添加额外信息
    怎样将Excel中的日期格式转换为文本格式
    为Tomcat页面设置访问权限(HTTP)
    为tomcat页面设置访问权限(BASIC认证)
    离线安装Maven FOR Eclipse插件
    sharepoint 获得上级和部门的封装函数
    SharePoint定制开发个性皮肤
  • 原文地址:https://www.cnblogs.com/soTired/p/4702871.html
Copyright © 2011-2022 走看看