DFS补题 红与黑 气候变暖

红与黑 链接： ACWing 1113

#include <bits/stdc++.h>

using namespace std;

const int N = 200;

int dx[5] = {1,-1,0,0} , dy[5] = {0,0,-1,1};
int h, w;
char g[N][N];

int dfs(int a, int b){
	g[a][b] = '#';
    int ans = 1;
	for (int i = 0; i < 4; i ++){
		int x = a + dx[i], y = b + dy[i];
		if (x >= 0 && x < h && y >= 0 && y < w && g[x][y] == '.'){
		    ans += dfs(x,y);
		}
	}
	return ans;
}

int main(){
	cin >> w >> h;
	while (w && h){	
		int a, b;
		for (int i = 0; i < h; i ++)
			for (int j = 0; j < w; j ++){
				cin >> g[i][j];
				if (g[i][j] == '@'){
					a = i;
					b = j;
				}	
			}
		cout << dfs(a,b) << endl;
		cin >> w >> h;
	}
	return 0;	
}

气候变暖 链接: AWing 1233

# include <bits/stdc++.h>

using namespace std;

const int N = 1010;

char g[N][N];
bool st[N][N];
int total, brink;
int n;
int dx[5] = {1,-1,0,0},dy[5] = {0,0,-1,1};

void dfs(int x , int y){
	st[x][y] = true;
	total ++;
	
	for (int i = 0; i < 4; i ++)
		if (g[x + dx[i]][y + dy[i]] == '.'){
				brink ++;
				break;
		}
	
	for (int i = 0; i < 4; i ++){
		int a = x + dx[i] , b = y + dy[i];
		if(a < 0 || a >= n || b < 0 || b >= n) continue;
		if (g[a][b] == '#' && !st[a][b]) dfs(a,b);
	}
	
}

int main(){
	cin >> n;
	for (int i = 0; i < n; i ++) cin >> g[i];
	int ans = 0;
	for (int i = 0; i < n; i ++)
		for (int j = 0; j < n; j ++){
			if (g[i][j] == '#' && !st[i][j]){
				total = brink = 0;
				dfs(i,j);
				if (total == brink) ans ++;
			}
		}
	cout << ans;
	return 0;
}