3620 Avoid The Lakes【dfs】

Avoid The Lakes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6947		Accepted: 3691

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：

给出三个数，前两个限定的是一个区域的宽和高，也就是矩阵的行和列，后一个代表有几个特殊点，后面对应的是点的坐标，规定，前后左右相邻的特殊点，算是一个集合，求出在这个范围内，元素最多的集合内有多少元素...

这个题是比较简单的 dfs 题目，只要用递归多次调用，加上特殊处理，以及统计个数的条件，不难求出答案，主要是理解 dfs 的工作原理...

#include<stdio.h>
#include<string.h>
int x[105][105],n,m,t,sum,max;
void dfs(int a,int b)
{
if(a<0||a>n||b<0||b>m)//越界
{
return;
}
if(x[a][b]==0)//不是需要的点
{
return;
}
++sum;//在运行的时候统计....
x[a][b]=0;
dfs(a+1,b);//对前后左右进行搜索判断
dfs(a-1,b);
dfs(a,b+1);
dfs(a,b-1);
}
int main()
{
int i,j,a,b;
while(~scanf("%d%d%d",&n,&m,&t))
{
memset(x,0,sizeof(x));
for(i=0;i<t;++i)
{
scanf("%d%d",&a,&b);
x[a][b]=1;//标记在二维数组上
}
max=sum=0;
for(i=0;i<=n;++i)
{
for(j=0;j<=m;++j)
{
if(x[i][j]==1)//查找和运算
{
dfs(i,j);
if(sum>max)
{
max=sum;//更新最大值
}
sum=0;
}
}
}
printf("%d\n",max);//输出
}
return 0;
}