哈尔滨理工大学软件与微电子学院程序设计竞赛——H.Maze【BFS & 并查集】

题目传送门

---

---

题解

• 不考虑 $Q$ 次询问的话，就是简单的 $BFS$
• 那么有了 $Q$ 次询问，那么直接使用 $并查集$ 维护即可
• 注意数组大小

---

AC-Code

#include <bits/stdc++.h>
using namespace std;

const int maxn = 3e3 + 7;
int N, M, Q;
struct NODE {
	int x, y;
	NODE(int x = 0, int y = 0) :x(x), y(y) {}
};
char mp[maxn][maxn];
bool vis[maxn][maxn];
int dx[] = { 0,0,1,-1 };
int dy[] = { 1,-1,0,0 };
int getId(int x, int y) {
	return x * M + y;
}
struct UFS {
	int f[maxn * maxn];
	unsigned int rank[maxn * maxn];
	UFS() {}
	void init() {
		for (int i = 0; i < maxn * maxn; ++i) {
			f[i] = i;	rank[i] = 0;
		}
	}
	int getF(int x) {
		return f[x] == x ? x : (f[x] = getF(f[x]));
	}
	void merge(int x, int y) {
		x = getF(x), y = getF(y);
		if (x == y)	return;
		if (rank[x] < rank[y])	f[x] = y;
		else	f[y] = x;
		if (rank[x] == rank[y])	++rank[x]; // 走else合并
	}
	bool isUnion(int x, int y) {
		return getF(x) == getF(y);
	}
}UF;
bool judge(int x, int y, int nx, int ny) {
	if (nx < 0 || nx >= N || ny < 0 || ny >= M)	return false;
	if (mp[x][y] == mp[nx][ny])	return false;
	return !vis[nx][ny];
}
void bfs(NODE start) {
	queue<NODE>q;
	q.push(start);
	vis[start.x][start.y] = true;
	while (!q.empty()) {
		NODE now = q.front();
		q.pop();
		for (int i = 0; i < 4; ++i) {
			int nx = now.x + dx[i];
			int ny = now.y + dy[i];
			if (judge(now.x, now.y, nx, ny)) {
				q.push(NODE(nx, ny));
				vis[nx][ny] = true;
				UF.merge(getId(nx, ny), getId(now.x, now.y));
			}
		}
	}
}
int ans[maxn * maxn];
int main() {
	while (cin >> N >> M >> Q) {
		memset(ans, 0, sizeof(ans));
		memset(vis, false, sizeof(vis));
		UF.init();
		for (int i = 0; i < N; ++i)
			for (int j = 0; j < M; ++j)
				cin >> mp[i][j];
		for (int i = 0; i < N; ++i) {
			for (int j = 0; j < M; ++j) {
				if (UF.getF(getId(i, j)) == getId(i, j)) {
					bfs(NODE(i, j));
				}
			}
		}
		for (int i = 0; i < N; ++i) {
			for (int j = 0; j < M; ++j) {
				++ans[UF.getF(getId(i, j))];
			}
		}
		for (int i = 0; i < Q; ++i) {
			int x, y;	cin >> x >> y;
			cout << ans[UF.getF(getId(x - 1, y - 1))] << endl;
		}
	}
}