『题解』SPOJ 12005 Grass Planting

模板题，难点在于怎样将点权变为边权。

我们可以考虑这样的事情，对于树上除了根之外的所有点，都有入边仅为 \(1\)， 出边可能为多条 则我们可以将这条入边的边权转化到这个点上。

树链剖分之后得到区间，我们可以知道对于树询问路径最后一段询问中 \(u,v\) 所在的重链为同一条，所以在转化后这段区间是连续的，我们只需要把这棵子树的根节点去掉即可， 即 \((id[v] + 1, id[u])\)。

Code

#include <bits/stdc++.h>

#define L u << 1
#define R u << 1 | 1

using namespace std;

const int maxn = 100010, maxm = maxn * 2;

int n, m;
int h[maxn], e[maxm], ne[maxm], idx; 
int dep[maxn], fa[maxn], sz[maxn], son[maxn];
int id[maxn], top[maxn], cnt;

struct Node
{
	int l, r;
	int sum, add;
}tr[maxn << 2];

void add(int a, int b)
{
	e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ; 
}

void dfs1(int u, int father, int depth)
{
	dep[u] = depth, fa[u] = father, sz[u] = 1;
	for (int i = h[u]; ~i; i = ne[i])
	{
		int j = e[i];
		if (j == father) continue;
		dfs1(j, u, depth + 1);
		sz[u] += sz[j];
		if (sz[j] > sz[son[u]]) son[u] = j;
	}
}

void dfs2(int u, int t)
{
	id[u] = ++ cnt, top[u] = t;
	if (!son[u]) return;
	dfs2(son[u], t);
	for (int i = h[u]; ~i; i = ne[i])
	{
		int j = e[i];
		if (j == fa[u] || j == son[u]) continue;
		dfs2(j, j);
	}
}

void pushup(int u)
{
	tr[u].sum = tr[L].sum + tr[R].sum;
}

void pushdown(int u)
{
	auto &root = tr[u], &left = tr[L], &right = tr[R];
	if (root.add)
	{
		left.add += root.add, left.sum += (left.r - left.l + 1) * root.add;
		right.add += root.add, right.sum += (right.r - right.l + 1) * root.add;
		root.add = 0;
	}
}

void build(int u, int l, int r)
{
	tr[u] = {l, r, 0, 0};
	if (l == r) return;
	int mid = l + r >> 1;
	build(L, l, mid), build(R, mid + 1, r);
	pushup(u);
}

void update(int u, int l, int r)
{
	if (tr[u].l >= l && tr[u].r <= r)
	{
		tr[u].sum += tr[u].r - tr[u].l + 1;
		tr[u].add ++ ;
		return;
	}
	pushdown(u);
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid) update(L, l, r);
	if (r > mid) update(R, l, r);
	pushup(u);
}

int query(int u, int l, int r)
{
	if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum;
	pushdown(u);
	int res = 0;
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid) res += query(L, l, r);
	if (r > mid) res += query(R, l, r);
	return res;
}

void update_path(int u, int v)
{
	while (top[u] != top[v])
	{
		if (dep[top[u]] < dep[top[v]]) swap(u, v);
		update(1, id[top[u]], id[u]);
		u = fa[top[u]];
	}
	if (dep[u] < dep[v]) swap(u, v);
	update(1, id[v] + 1, id[u]);
}

int query_path(int u, int v)
{
	int res = 0;
	while (top[u] != top[v])
	{
		if (dep[top[u]] < dep[top[v]]) swap(u, v);
		res += query(1, id[top[u]], id[u]);
		u = fa[top[u]];
	}
	if (dep[u] < dep[v]) swap(u, v);
	res += query(1, id[v] + 1, id[u]);
	return res;
}

int main()
{
	scanf("%d%d", &n, &m);
	
	memset(h, -1, sizeof h);
	for (int i = 0; i < n - 1; i ++ )
	{
		int a, b;
		scanf("%d%d", &a, &b);
		add(a, b), add(b, a);
	}
	
	dfs1(1, -1, 1);
	dfs2(1, 1);
	build(1, 1, n);
	
	while (m -- )
	{
		char op[2];
		int a, b;
		scanf("%s%d%d", op, &a, &b);
		if (*op == 'P') update_path(a, b);
		else printf("%d\n", query_path(a, b));
	}
	
	return 0;
}