POJ3620-Avoid The Lakes-DFS-优快云博客

这篇博客探讨了在最近的洪水中，Farmer John的农场被水淹没的情况。由于他的奶牛极度害怕水，问题变得更加严重。文章介绍了一个算法，该算法用于确定农场上最大的‘湖’（被水淹没的区域）的大小，以便根据湖泊面积来计算保险赔偿。输入包括农场的行数、列数和被水淹没的单元格数量，输出为最大湖泊包含的单元格数。代码示例展示了如何使用深度优先搜索来解决这个问题。

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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<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,m,k,a[101][101],s;
void dfs(int x,int y)
{
	int i,j,tx,ty,e[4][2]={1,0,0,1,-1,0,0,-1};//方向
	if(x<1||y<1||x>n||y>m||a[x][y]==0)//越界或者没有
	return;
	a[x][y]=0;//标记
	s++;
	for(i=0;i<4;i++)
	dfs(x+e[i][0],y+e[i][1]);//遍历
}
int main()
{
	int i,j,t,x,y,sum;
	while(scanf("%d%d%d",&n,&m,&k)!=EOF)
	{
		sum=0;
		memset(a,0,sizeof(a));
	for(i=1;i<=k;i++)
	{
		scanf("%d%d",&x,&y);
		a[x][y]=1;//标记
	}
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
	    if(a[i][j]==1)
	    {
	    	s=0;
	    	dfs(i,j);//遍历
	    	sum=max(s,sum);//判断
		}
	}	
	printf("%d\n",sum);//输出
   }
}