Avoid The Lakes
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 這道搜索題，搜了我好久就沒出正確結果。看着別人的AC代碼，也沒找出錯誤。找了好久才發現自己的數組下標從零開始出問題了。哎！以後還是少從零開始吧。
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<stack>
#define MAX(x,y) x>y?x:y;
#define MIN(x,y) x<y?x:y;
using namespace std;
int map[110][110];
int mark[110][110];
int n,m,k,key;
int dfs(int x,int y)
{
int i;
const int dir[4][2]={-1,0,1,0,0,1,0,-1};
for(i=0;i<4;i++)
{
int xx=x+dir[i][0];
int yy=y+dir[i][1];
if(xx>=1&&yy>=1&&xx<=n&&yy<=m&&map[xx][yy]==1&&!mark[xx][yy])
{
mark[xx][yy]=1;
key++;
dfs(xx,yy);
}
}
return key;
}
int main()
{
int i,j,x,y,ans;
while(cin>>n>>m>>k)
{
ans=0;
memset(map,0,sizeof(map));
memset(mark,0,sizeof(mark));
for(i=0;i<k;i++)
{
cin>>x>>y;
map[x][y]=1;
}
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(map[i][j]==1&&!mark[i][j])
{
key=1;
mark[i][j]=1;
ans=MAX(ans,dfs(i,j));
}
}
}
cout<<ans<<endl;
}
return 0;
}