BZOJ3631

BZOJ 3631

• 题目

  BZOJ3631

• 分析

  考虑 $LCA$ + 差分

  主要算法不讲了。。

  现在考虑一下细节：$a_1\to a_2,a_2\to a_3,...a_{n-1}\to a_n$ 标记点时在 $a_1\to a_2,a_2\to a_3$,$a_2$ 被标记了两次，最后一个点时餐厅不用放糖果，所以这些点的最后的数目都要减一。

• 代码

  const int SIZE = 300010;
  int f[SIZE][20], d[SIZE], dist[SIZE];
  int ver[2 * SIZE], Next[2 * SIZE], edge[2 * SIZE], head[SIZE];
  int T, n, m, tot, t;
  queue<int> q;
  int a[SIZE];
  int cnt[SIZE];
  int vis[SIZE];
  void add(int x, int y, int z) {
         
  	ver[++tot] = y; edge[tot] = z; Next[tot] = head[x]; head[x] = tot;
  }
  
  void bfs() {
         
  	q.push(1); d[1] = 1;
  	while (q.size()) {
         
  		int x = q.front(); q.pop();
  		for (int i = head[x]; i; i = Next[i]) {
         
  			int y = ver[i];
  			if (d[y]) continue;
  			d[y] = d[x] + 1;
  			dist[y] = dist[x] + edge[i];
  			f[y][0] = x;
  			for (int j = 1; j <= t; j++)
  				f[y][j] = f[f[y][j - 1]][j - 1];
  			q.push(y);
  		}
  	}
  }
  
  int lca(int x, int y) {
         
  	if (d[x] > d[y]) swap(x, y);
  	for (int i = t; i >= 0; i--)
  		if (d[f[y][i]] >= d[x]) y = f[y][i];
  	if (x == y) return x;
  	for (int i = t; i >= 0; i--)
  		if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  	return f[x][0];
  }
  
  void dfs(int x)
  {
         
  	vis[x] = 1;
  	for (int i = head[x]; i; i = Next[i])
  	{
         
  		int y = ver[i];
  		if (vis[y]) continue;
  		dfs(y);
  		cnt[x] += cnt[y];
  	}
  }
  
  int main() {
         
  	read(n);
  	t = (int)(log(n) / log(2)) + 1;
  	for (int i = 1; i <= n; i++) head[i] = d[i] = 0;
  	tot = 0;
  	for (int i = 1; i <= n; i++) read(a[i]);
  	for (int i = 1; i < n; i++) {
         
  		int x, y;
  		read(x);
  		read(y);
  		add(x, y, 1), add(y, x, 1);
  	}
  	bfs();
  	for (int i = 1; i <= n - 1; i++)
  	{
         
  		int s = a[i];
  		int t = a[i + 1];
  		int _ = lca(s,t);
  		cnt[s] ++;
  		cnt[t] ++;
  		cnt[_]--;
  		cnt[f[_][0]]--;
  	}
  	dfs(1);
      for(int i = 2;i <= n;i++) cnt[a[i]]--;
      for(int i = 1;i <= n;i++) 
      {
         
      	out(cnt[i]);
      	puts("");
      }
  	return 0;
  }
  
  

• 题型

  $LCA$ + 差分