POJ Problem 3620 Avoid The Lakes 【DFS】

Avoid The Lakes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7748		Accepted: 4092

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

Source

USACO 2007 November Bronze

用DFS从左上角开搜，如果碰到湖水，就把他填平。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
bool map[105][105];

int n, m, k;
int dfs(int x, int y) {
    if (!map[x][y] || x < 1|| y < 1 || x > n || y > m) return 0;
    else {
        map[x][y] = false;
        return 1+dfs(x, y+1)+dfs(x, y-1)+dfs(x+1, y)+dfs(x-1, y);
    }
}
int main() {
    int row, col;
    while (scanf("%d%d%d", &n, &m, &k) != EOF) {
        memset(map, false, sizeof(map));
        int ans = 0;
        for (int i = 0; i < k; i++) {
            scanf("%d%d", &row, &col);
            map[row][col] = true;
        }
        //进行从左上角的遍历，ans统计湖面积最大值。
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                ans = max(ans, dfs(i, j));
            }
        }
        printf("%d
", ans);
    }
    return 0;
}