调题的时候先看看这个吧

大锑问题之
1.树上问题,先将子树答案更新后,用新的值更新答案

//可并堆合并
struct MHN{ ll val, lc, rc, sz, dist, sum;} H[100007];
int merge(int x, int y)
{
	if((!x) || (!y)) return x | y;
	if(H[x].val < H[y].val) swap(x, y);
	H[x].sz += H[y].sz;//H[y]会改掉,应该先加完再合并, 100pts
	H[x].sum += H[y].sum;
	H[x].rc = merge(H[x].rc, y);
	if(H[H[x].lc].dist < H[H[x].rc].dist) swap(H[x].lc, H[x].rc);
	H[x].dist = H[H[x].rc].dist + 1;
//	H[x].sz += H[y].sz;//H[y]被改掉, 8pts
//	H[x].sum += H[y].sum;
//  也可以写成这样:
//	H[x].sz = H[H[x].lc].sz + H[H[x].rc].sz + 1;
//	H[x].sum = H[H[x].lc].sum + H[H[x].rc].sum + H[x].val;
	return x;
}