poj3468 A Simple Problem with Integers

普通序列，区间更新，区间查询。

区间更新：

每个结点的sum信息，表示以这个结点为根的所有子树结点的和，包括根结点。

void add(int a, int b, int c) {
  if(a > b) swap(a, b);
  Node *left, *mid, *right, *o;
  split(ss.root, a, left, o);
  split(o, b-a+1, mid, right);
  mid->v += c;
  mid->sum += (i64)c*mid->s;
  mid->lazy += c;
  ss.root = merge(merge(left, mid), right);
}

 区间求和：

这个题需要用i64，v与延迟标记lazy，以及sum都应该是i64类型。。

void query_sum(int a, int b) {
  Node *left, *mid, *right, *o;
  split(ss.root, a, left, o);
  split(o, b-a+1, mid, right);
  printf ( "%lld\n", mid->sum );
  ss.root = merge(merge(left, mid), right);
}

 

maintain()和pushdown():

特别注意一下sum。。

  void maintain() {
    s = ch[0]->s + ch[1]->s + 1;
    sum = v;
    if ( ch[0] != null ) sum += ch[0]->sum;
    if ( ch[1] != null ) sum += ch[1]->sum;
  }
  void pushdown() {
    if(lazy) {
      ch[0]->v += lazy;
      ch[0]->sum += lazy*ch[0]->s;
      ch[0]->lazy += lazy;
      ch[1]->v += lazy;
      ch[1]->sum += lazy*ch[1]->s;
      ch[1]->lazy += lazy;
    }
    lazy = 0;
  }