BFS流程步骤：

1.首结点入队列并标记

2.搜寻首结点的上，下，左，右结点，满足条件就入队列并标记

3.首结点出队列，下一个结点成为首节点，计数器加一

4.重复2，3步骤，直到队列中全部的结点出队列

BFS

#include <bits/stdc++.h>
#include <queue>
using  namespace std;
int n, m, x, y, ans, cnt = 0;
const int maxn = 1005;
int area[maxn][maxn];
int X[4] = {-1, 1, 0, 0}; //上下左右四个方向 
int Y[4] = {0, 0, -1, 1};
bool F[maxn][maxn] = {false};

struct node {
	int x;
	int y;
}Node, top;

bool judge(int x, int y) {
	if(x < 0 || y < 0 || x >= n || y >= m) //越界不入队列 
	return false;
	if(F[x][y] == true || area[x][y] == 0) //被选过了或者没有树不入队列 
	return false; 
	return true;
}

void BFS(int x, int y) {
	queue<node> q;
	Node.x = x;
	Node.y = y;
	q.push(Node);
	while (!q.empty()) {
		top = q.front();
		int lx = top.x;
		int ly = top.y;
		for (int i = 0; i < 4; i++) {
			if (judge(lx + X[i], ly + Y[i]))
			{
				Node.x = lx + X[i];
				Node.y = ly + Y[i]; 
				q.push(Node);
				F[Node.x][Node.y] = true;
			}
		}
		ans++;
		F[lx][ly] = true;
		q.pop();
	}
}

int main() {
	cin >> n >> m >> x >> y;
	for (int i = 0; i < n; i ++) {
		for (int j = 0; j < m; j++) {
			cin >> area[i][j];
			if (area[i][j] == 1) cnt++;
		}
	}
	
	if (area[x][y] == 1) {
		BFS(x, y);
	}
	cout << ans << endl;
	double rate = (double)ans / cnt;
	if (rate >= 0.2)
	cout << "获得成就：生 态 灭 绝 者";
	else if (rate >= 0.1)
	cout << "获得成就：伐伐伐伐伐木工";
	else 
	cout <<"获得成就：就这？"; 	
		
	
	return 0;
} 

DFS

#include<stdio.h>
#include<iostream>
using namespace std;
const int N=1005;
int n,m,k,fx,fy,ex,ey,sx,sy,ans,sum;
double rate;
int ma[N][N],mas[N][N];
int opx[5]={1,-1,0,0};
int opy[5]={0,0,-1,1};

void dfs(int x,int y)
{
		int nx,ny;
		for(int i=0;i<4;i++)
		{
			nx=x+opx[i];
			ny=y+opy[i];
			if(nx>=0&&nx<=n-1&&ny>=0&&ny<=m-1&&mas[nx][ny]==0&&ma[nx][ny]==1)
			{
				ans++;
				mas[nx][ny]=1;
				dfs(nx,ny);
			}
		}
	
}
int main()
{
	scanf("%d %d",&n,&m);
	scanf("%d %d",&sx,&sy);
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			scanf("%d",&ma[i][j]);
			if(ma[i][j]==1) sum++;
		}
	}
	if(ma[sx][sy]==1) 
	{
        ans++;
	    mas[sx][sy]=1;
	    dfs(sx,sy);
	}

	rate=(double)ans/sum;
	if(rate>=0.2)
	{
		printf("%d\n",ans);
		printf("获得成就：生 态 灭 绝 者\n");
	}
	else if(rate<0.2&&rate>=0.1)
	{
		printf("%d\n",ans);
		printf("获得成就：伐伐伐伐伐木工\n");
	}
	else
	{
		printf("%d\n",ans);
		printf("获得成就：就这？\n");
	}
	return 0;
}