AcWing 1113. 红与黑

题目传送门

一、dfs+Flood Fill

#include 
using namespace std;

const int N = 25;
// dfs 实现flood fill 算法

int n, m;
char g[N][N];
bool st[N][N];

int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};

int dfs(int x, int y) {
    // x,y贡献了1个结点数
    int cnt = 1;
    //老师喜欢到了再判定，我喜欢判定再决定是不是可以走
    st[x][y] = true; // Flood Fill 不用回溯

    for (int i = 0; i < 4; i++) {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (g[a][b] != '.') continue;
        if (st[a][b]) continue;
        //累加子孙后代的贡献数
        cnt += dfs(a, b);
    }
    //返回此子树的节点个数
    return cnt;
}

int main() {
    while (cin >> m >> n, n || m) { //先输入列数，再输入行数，小坑
        for (int i = 0; i < n; i++) cin >> g[i];

        //找出起点
        int x, y;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (g[i][j] == '@') {
                    x = i;
                    y = j;
                }

        //多组测试数据
        memset(st, 0, sizeof st);
        cout << dfs(x, y) << endl;
    }

    return 0;
}

二、bfs+Flood Fill

#include 
using namespace std;
#define x first
#define y second
typedef pair PII;

const int N = 30;

char g[N][N];
int n, m;
int dx[] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
bool st[N][N];

int bfs(int x, int y) {
    int cnt = 1;
    queue q;
    q.push({x, y});

    while (q.size()) {
        PII t = q.front();
        q.pop();
        int x = t.x, y = t.y;
        for (int i = 0; i < 4; i++) {
            int a = x + dx[i], b = y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (st[a][b]) continue;
            if (g[a][b] != '.') continue;
            st[a][b] = true;
            q.push({a, b});
            cnt++;
        }
    }
    return cnt;
}

int main() {
    while (cin >> m >> n, n || m) {
        memset(st, 0, sizeof st);
        for (int i = 0; i < n; i++) cin >> g[i];

        int x, y;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++)
                if (g[i][j] == '@')
                    x = i, y = j;
        }
        cout << bfs(x, y) << endl;
    }
    return 0;
}