poj3620Avoid The Lakes

本文介绍了一个名为AvoidTheLakes的问题,这是一个经典的图遍历问题。在一个由N行M列组成的矩形网格中,有K个单元格被水淹没。文章通过深度优先搜索(DFS)算法来确定最大的‘湖泊’覆盖了多少个单元格,并提供了完整的C语言实现代码。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Avoid The Lakes
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

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
#include<stdio.h>
#include<string.h>
int map[110][110];
int n,m;
int ans;
void dfs(int x,int y)
{
	if(map[x][y]==0)
	return;
	else if(map[x][y]==1)
	{
	map[x][y]=0;
	ans++;
	dfs(x-1,y);
	dfs(x+1,y);
	dfs(x,y+1);
	dfs(x,y-1);
    }
}
int main()
{
	int u,v,k,i,j;
	int a[10010],b[10010];
	while(~scanf("%d%d%d",&n,&m,&k))
	{
		memset(map,0,sizeof(map));
		while(k--)
		{
			scanf("%d%d",&u,&v);
			map[u][v]=1;
		}
		int max=0;
		for(i=1;i<=n;i++)
		for(j=1;j<=m;j++)
		{
		if(map[i][j])
		{
			ans=0;
			dfs(i,j);
		}
			if(max<ans)
			max=ans;
		}
		printf("%d\n",max);
	}
	return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值