POJ - 3620-Avoid The Lakes 【DFS】_farmer john's farm was flooded in the most recent -优快云博客

本文链接：https://blog.youkuaiyun.com/Li_Hongcheng/article/details/79338056

本文介绍了一个名为AvoidTheLakesPOJ-3620的问题，该问题是关于在一个由N行M列组成的矩形网格中找到最大的湖泊（连续水区域）。湖泊是由相邻的水格子组成的，相邻是指两个格子共享一个边。文章提供了一个C++程序示例来解决这个问题，使用深度优先搜索算法遍历网格并计算最大湖泊的大小。

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Avoid The Lakes POJ - 3620
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: 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

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[105][105];
int ans,n,m;
int cmp(int a,int b)
{
    return a<b;
}


void dfs(int x,int y)
{
    if(a[x][y]==1||x<1||x>n||y<1||y>m)
        return;
    a[x][y]=1;
    ans++;

    dfs(x-1,y);
    dfs(x+1,y);
    dfs(x,y+1);
    dfs(x,y-1);
}


int main()
{
    int k;
    int x,y;
    while(~scanf("%d%d%d",&n,&m,&k))
    {

        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
                a[i][j]=1;
        }
        while(k--)
        {
            scanf("%d%d",&x,&y);
            a[x][y]=0;
        }
        int max=-1;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {

                if(a[i][j]==0)
                {
                    ans=0;
                    dfs(i,j);
                    if(max<ans)
                        max=ans;
                }
            }
        }
        printf("%d\n",max);
    }
}