POJ 3620--avoid the Lakes "DFS"

Avoid the Lakes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6775		Accepted: 3620

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

Test instructions: The first line inputs N, M, K. Representing an area of n * M size, K represents a K-puddle, and the next K-row represents the coordinates of a K-puddle,

A puddle can form a pit of water and water around it, asking how many puddles are in the pits.

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include < Queue>using namespace Std;int Dir[4][2] = {1, 0,-1, 0, 0,-1, 0, 1};int map[110][110];int vis[110][110];int N, M, K, a    Ns;bool Check (int x, int y) {if (x < 1 | | x > N | | y < 1 | | y > m) return 0;    if (map[x][y] = = 0) return 0;    if (vis[x][y] = = 1) return 0; return 1;}    void Dfs (int x,int y) {ans++;    Vis[x][y] = 1;        for (int i = 0; i < 4; ++i) {int FX = x + dir[i][0];        int fy = y + dir[i][1];    if (check (FX, FY)) DFS (FX, FY);        }}int Main () {while (scanf ("%d%d%d", &n, &m, &k)! = EOF) {memset (map, 0, sizeof (map));            while (k--) {int x, y;            scanf ("%d%d", &x, &y);        Map[x][y] = 1;        } memset (Vis, 0, sizeof (VIS));        int sum = 0;  for (int i = 1;i <= N, ++i) for (int j = 1; j <= m; ++j) {ans = 0;          if (Map[i][j]) {DFS (i, j);            sum = max (sum, ans);    }} printf ("%d\n", sum); } return 0;}

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POJ 3620--avoid the Lakes "DFS"