POJ 3620 Avoid The Lakes (for the longest connecting line) (DFS)

Description

Farmer John's farm was flooded in the most recent storm, a fact-only aggravated by the information so his cows is death Ly afraid of water. His insurance agency would only repay him, however, a 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 ≤ ) columns. Each cell in the grid is either dry or submerged, and exactly K (1≤ k ≤ N x M) of the Cells are submerged. As one would expect, a lake has a central cell to which and other cells connect by sharing a long edge (not a corner). Any cell this shares a long edge with the central cell or shares a long edge with any connected cell becomes a connected C Ell 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 II space separated integers that is its row and co Lumn: R and C

Output

* Line 1:the number of cells the the largest lake contains.

Sample Input

3 4 53 22 23 12 31 1

Sample Output

4
Give a string of coordinates, one for each of them, and calculate the maximum number of coordinates that each consists of.

1#include <cstdio>2#include <string.h>3 intdx[4]={-1,1,0,0};4 intdy[4]={0,0,-1,1};5 intn,m,map[ Max][ Max],x,y,i,max,ans,k,j;6 voidFintXinty)7 {8     intI,nx,ny;9      for(i =0; I <4; i++)Ten     { Onenx=x+Dx[i]; Any=y+Dy[i]; -         if(NX >0&& NY >0&& NX <= N && ny <=m && map[nx][ny] = =1) -         { theans++; -map[nx][ny]=0; - f (nx,ny); -         } +     } - } + intMain () A { at      while(SCANF (" %d%d%d", &n,&m,&k)! =EOF) -     { -max=0; -memset (Map,0,sizeof(map)); -          for(i =0; I < K; i++) -         { inscanf"%d%d",&x,&y); -map[x][y]=1;//1 means not passing . to         } +          for(i =1; I <=n; i++) -         { the              for(j =1; J <= M; J + +) *             { $                 Panax Notoginseng                 if(Map[i][j] = =1) -                 {     theans=1; +map[i][j]=0; A f (i,j); theMax=max>ans?Max:ans; +                 } -                  $             } $              -         } -printf"%d\n", max); the     } -}

POJ 3620 Avoid The Lakes (for the longest connecting line) (DFS)