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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: 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
题意：给出一个矩阵，其中有些格子干燥、有些潮湿。
如果一个潮湿的格子的相邻的四个方向有格子也是潮湿的，那么它们就可以构成更大
的湖泊，求最大的湖泊。
也就是求出最大的连在一块儿的潮湿的格子的数目。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define MAX 110
#define inf 0x7ffffff
int map[MAX][MAX];
int vis[MAX][MAX];
int fx[4]={1,0,-1,0};
int fy[4]={0,1,0,-1};
int n,m,cnt;
int valid(int i,int j)
{
if(i>0&&i<=n&&j>0&&j<=m)
return 1;
else
return 0;
}
void DFS(int i,int j)
{
if(!valid(i,j))
return ;
map[i][j]=0;
for(int k=0;k<4;k++)
{
int x=i+fx[k];
int y=j+fy[k];
if(valid(x,y)&&map[x][y])
{
cnt++;
DFS(x,y);
}
}
}
int main()
{
int k;
while(~scanf("%d%d%d",&n,&m,&k))
{
memset(map,0,sizeof(map));
for(int i=1;i<=k;i++)
{
int a,b;
scanf("%d%d",&a,&b);
map[a][b]=1;
}
int ans=-inf,t=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
cnt=1;
if(map[i][j])
DFS(i,j);
ans=max(ans,cnt);
}
printf("%d\n",ans);
}
return 0;
}
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原文地址：http://blog.csdn.net/lh__huahuan/article/details/47354787
