Avoid The Lakes(DFS)

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 Avoid The Lakes
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu
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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: 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 53 22 23 12 31 1

Sample Output

4
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 110;bool map[maxn][maxn];int dx[4]={0,1,-1,0};int dy[4]={1,0,0,-1};//int i,j;    // 这里定义了全局变量i,j;但由于i,j在主函数main与自定义dfs函数都有用到 ;因此此处不能全局定义;;;; int n,m,k,ans,sum;void dfs(int x,int y){sum++;if(!map[x][y]) return;map[x][y]=false;for(int i=0;i<4;i++){int nx=x+dx[i];int ny=y+dy[i];if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&map[nx][ny])dfs(nx,ny);}}int main(){while(scanf("%d %d %d",&n,&m,&k)!=EOF){ans=0;memset(map,false,sizeof(map));for(int i=1;i<=k;i++){int x,y;scanf("%d %d",&x,&y);map[x][y]=true;}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(map[i][j]){sum=0;dfs(i,j);ans=max(ans,sum);}}}printf("%d\n",ans);}return 0;}// 注意全局变量和局部变量


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