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思路分析:

    考虑使用BFS, 扩展队头结点时将队头结点的上, 下, 左, 右, 左上, 左下, 右上, 右下未被访问的草地入队即可, 具体实现如下AC代码所示:

//CH2907_乳草的入侵
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
const int MAX = 105, INF = 0x3f3f3f3f;
const int dx[8] = {0, -1, -1, -1, 0, 1, 1, 1}, dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
char G[MAX][MAX]; int R, C, mx, my;
int d[MAX][MAX];//d[i][j]: 位置i行j列处的草地被占领的星期数 
int main(){
	int tx, ty; scanf("%d %d %d %d", &C, &R, &tx, &ty), mx = R - ty + 1, my = tx;
	for(int i = 1; i <= R; ++i) scanf("%s", G[i] + 1);
	memset(d, 0x3f, sizeof(d)), d[mx][my] = 0; queue<pii> qu; qu.push(mp(mx, my));
	while(!qu.empty()){
		int x = qu.front().fi, y = qu.front().se; qu.pop();
		for(int i = 0; i <= 7; ++i){
			int a = x + dx[i], b = y + dy[i];
			if(a >= 1 && a <= R && b >= 1 && b <= C && G[a][b] == '.' && d[a][b] == INF)
				d[a][b] = d[x][y] + 1, qu.push(mp(a, b));
		}
	} 
	int res = -INF; 
	for(int i = 1; i <= R; ++i)
		for(int j = 1; j <= C; ++j) if(d[i][j] != INF) res = max(res, d[i][j]);
	cout << res << endl;
	return 0;
}