POJ 1979: Red and Black

題目在此

解題思路：直接 DFS 或 BFS 就好了。

以前被 STL 拖事後腿，偏執勁兒又上來了，此次刻意不用 std::queue，本身寫隊列，BFS 不用遞歸，結果就是……代碼又臭又長，讓人不忍直視……

貼出來吧，就當是個教訓。

#include <cstdio>

const int N = 20;
char room[N][N];
int W, H;

struct point {
    int x, y;

    point() {}
    point(int r, int c) : x(r), y(c) {
    }
} queue[N *N];

int head, tail;

void enqueue(const point &p) {
    queue[tail++] = p;
}

point &dequeue() {
    return queue[head++];
}

void clear() {
    head = tail = 0;
}

bool isEmpty() {
    return head == tail;
}

int BFS() {
    int ans = 0;
    while (!isEmpty()) {
        point p = dequeue();
        ++ans;

        if (p.y + 1 < W && room[p.x][p.y + 1] != '#') {
            enqueue(point(p.x, p.y + 1));
            room[p.x][p.y + 1] = '#';
        }

        if (p.x + 1 < H && room[p.x + 1][p.y] != '#') {
            enqueue(point(p.x + 1, p.y));
            room[p.x + 1][p.y] = '#';
        }

        if (p.y - 1 >= 0  && room[p.x][p.y - 1] != '#') {
            enqueue(point(p.x, p.y - 1));
            room[p.x][p.y - 1] = '#';
        }

        if (p.x - 1 >= 0 && room[p.x - 1][p.y] != '#') {
            enqueue(point(p.x - 1, p.y));
            room[p.x - 1][p.y] = '#';
        }
    }

    return ans;
}

int main() {
    while (scanf("%d %d", &W, &H) && (W || H)) {
        for (int i = 0; i < H; ++i) {
            for (int j = 0; j < W; ++j) {
                scanf(" %c", &room[i][j]);
            }
        }

        for (int i = 0; i < H; ++i) {
            for (int j = 0; j < W; ++j) {
                if (room[i][j] == '@') {
                    enqueue(point(i, j));
                    room[i][j] = '#';
                    goto solve;
                }
            }
        }

solve:
        printf("%d\n", BFS());
        clear();
    }

    return 0;
}