POJ 3620 Avoid The Lakes

DFS的应用, 求最大连通分量。刚开始在想要不要从每个格子开始深搜，并不断更新最大值。后来参考了大神的博客之后发现并不需要复原，直接变成0就好。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int MAXN = 105;int maze[MAXN][MAXN];int dir[4][2] = {-1, 0, 1, 0, 0, 1, 0, -1};int ans, cnt, C, R, K;void dfs(int x, int y){if (x < 1 || x > R || y < 1 || y > C || maze[x][y] == 0)return;maze[x][y] = 0;cnt++;for (int i = 0; i < 4; i++)dfs(x + dir[i][0], y + dir[i][1]);}int main(){while (scanf("%d%d%d", &R, &C, &K) != EOF){for (int i = 1; i <= R; i++)for (int j = 1; j <= C; j++)maze[i][j] = 0;while (K--){int i, j;scanf("%d%d", &i, &j);maze[i][j] = 1;}ans = 0;for (int i = 1; i <= R; i++)for (int j = 1; j <= C; j++)if (maze[i][j]){cnt = 0;dfs(i, j);ans = max(cnt, ans);}printf("%d\n", ans);}}