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:
    Go断后,Dart冲前,Google的野心
    gcc dynamic load library
    Go http server 高并发
    还是Go 为了伟大的未来
    windows go dll 框架
    Go cookie
    Go web ajax project
    hdoj 2844 Coins
    hdoj 1203 I NEED A OFFER!
    hdoj 2546 饭卡
  • 原文地址:https://www.cnblogs.com/soTired/p/4702871.html
Copyright © 2011-2022 走看看