[Vani有约会]雨天的尾巴

题解

这题做法很多呢！
目前我只试过线段树合并，因为最近在学

对于每一个点开一颗关于粮食类型的线段树
维护数量和区间最多的类型
然后考虑到合并子树的过程就类似于树上差分后做前缀和
所以我们可以树上差分一下，省去树剖的一只 (log) （但求 ( ext {lca}) 还是打了树剖）
最后就是套路地合并过程了！！

(Code)

#include<cstdio>
#include<iostream>
using namespace std;

const int N = 1e5 + 5;
int n, m, Len, h[N], tot;

struct edge{int to, nxt;}e[N << 1];
inline void add(int u, int v){e[++tot] = edge{v, h[u]}, h[u] = tot;}

int top[N], fa[N], dep[N], siz[N], son[N];
void dfs1(int x)
{
	siz[x] = 1;
	for(register int i = h[x]; i; i = e[i].nxt)
	{
		int v = e[i].to;
		if (v == fa[x]) continue;
		fa[v] = x, dep[v] = dep[x] + 1, dfs1(v), siz[x] += siz[v];
		if (siz[v] > siz[son[x]]) son[x] = v;
	}
}
void dfs2(int x)
{
	if (son[x]) top[son[x]] = top[x], dfs2(son[x]);
	for(register int i = h[x]; i; i = e[i].nxt)
	{
		int v = e[i].to;
		if (v == fa[x] || v == son[x]) continue;
		top[v] = v, dfs2(v);
	}
}
inline int LCA(int x, int y)
{
	int fx = top[x], fy = top[y];
	while (fx ^ fy)
	{
		if (dep[fx] > dep[fy]) x = fa[fx], fx = top[x];
		else y = fa[fy], fy = top[y];
	}
	if (dep[x] < dep[y]) return x;
	return y;
}

struct ask{int x, y, z;}a[N];

int size, rt[N];
struct Tree{int sum, col, ls, rs;}seg[N << 6];
inline void pushup(int p)
{
	if (seg[seg[p].ls].sum >= seg[seg[p].rs].sum)
	{
		seg[p].sum = seg[seg[p].ls].sum;
		seg[p].col = seg[seg[p].ls].col;
	}
	else{
		seg[p].sum = seg[seg[p].rs].sum;
		seg[p].col = seg[seg[p].rs].col;
	}
}
void insert(int &p, int l, int r, int x, int v)
{
	if (!p) p = ++size;
	if (l == r)
	{
		seg[p].sum += v, seg[p].col = l;
		return;
	}
	int mid = (l + r) >> 1;
	if (x <= mid) insert(seg[p].ls, l, mid, x, v);
	else insert(seg[p].rs, mid + 1, r, x, v);
	pushup(p);
}
int merge(int x, int y, int l, int r)
{
	if (!x || !y) return x | y;
	if (l == r)
	{
		seg[x].sum += seg[y].sum, seg[x].col = l;
		return x;
	}
	int mid = (l + r) >> 1;
	seg[x].ls = merge(seg[x].ls, seg[y].ls, l, mid);
	seg[x].rs = merge(seg[x].rs, seg[y].rs, mid + 1, r);
	pushup(x); return x;
}

int ans[N];
void dfs(int x)
{
	for(register int i = h[x]; i; i = e[i].nxt)
	{
		int v = e[i].to;
		if (v == fa[x]) continue;
		dfs(v), rt[x] = merge(rt[x], rt[v], 1, Len);
	}
	if (seg[rt[x]].sum) ans[x] = seg[rt[x]].col;
}

int main()
{
	scanf("%d%d", &n, &m);
	int x, y, z;
	for(register int i = 1; i < n; i++) scanf("%d%d", &x, &y), add(x, y), add(y, x);
	dfs1(1), top[1] = 1, dfs2(1);
	for(register int i = 1; i <= m; i++) 
		scanf("%d%d%d", &x, &y, &z), a[i] = ask{x, y, z}, Len = max(Len, z);
	for(register int i = 1; i <= m; i++)
	{
		int lca = LCA(a[i].x, a[i].y);
		insert(rt[a[i].x], 1, Len, a[i].z, 1), insert(rt[a[i].y], 1, Len, a[i].z, 1);
		insert(rt[lca], 1, Len, a[i].z, -1);
		if (fa[lca]) insert(rt[fa[lca]], 1, Len, a[i].z, -1);
	}
	dfs(1);
	for(register int i = 1; i <= n; i++) printf("%d
", ans[i]);
}