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​​

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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$并查集$
• 注意数组大小

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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;
    }
  }
}