洛谷P3384 【模板】树链剖分 树剖+线段树

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洛谷P3384 【模板】树链剖分

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标签

• 树链剖分

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简明题意

• 树剖的模板题。给出一棵有根树，需要你支持四种操作
  1. 修改：让树上u、v这条路径的所有节点都增加c
  2. 查询：查询树上u、v这天路径的权值之和
  3. 修改：让树上u的子树(包含u)全部增加c
  4. 查询：查询子树u（包含u）的权值之和

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思路

• 模板题，思路略，仅提供几点注意事项

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注意事项

• 搞清楚两次dfs得到的数组的含义
• 在路径查询或修改时，注意比较的是节点所在链的顶端的深度，而不是节点的深度。如果不用节点所在链的顶端的深度，会导致重复统计，从而得不到正确结果

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总结

• 无

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AC代码

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

const int maxn = 1e5 + 10;

int n, m, root, mod, a[maxn];
vector<int> g[maxn];

int dep[maxn], fa[maxn], siz[maxn], son[maxn];
void dfs1(int u, int f, int deep)
{
   dep[u] = deep;
   fa[u] = f;
   siz[u] = 1;

   int max_son = -1;
   for (auto& v : g[u])
   	if (v != f)
   	{
   		dfs1(v, u, deep + 1);
   		siz[u] += siz[v];
   		if (siz[v] > max_son)
   			max_son = siz[v], son[u] = v;
   	}
}

int id[maxn], cnt, w[maxn], top[maxn];
void dfs2(int u, int topf)
{
   id[u] = ++cnt;
   w[cnt] = a[u];
   top[u] = topf;
   if (son[u])
   {
   	dfs2(son[u], topf);
   	for (auto& v : g[u])
   		if (v != fa[u] && v != son[u])
   			dfs2(v, v);
   }
}

struct Node
{
   int l, r, sum;
   int tag;
};

Node tree[maxn * 4];

void spread(int o)
{
   if (tree[o].tag)
   {
   	tree[o].sum += tree[o].tag * (tree[o].r - tree[o].l + 1);
   	if (tree[o].l != tree[o].r)
   		tree[o * 2].tag += tree[o].tag, tree[o * 2 + 1].tag += tree[o].tag;
   	tree[o].tag = 0;
   }
}

void update(int o)
{
   if (tree[o].l != tree[o].r)
   {
   	spread(o * 2), spread(o * 2 + 1);
   	tree[o].sum = (tree[o * 2].sum + tree[o * 2 + 1].sum) % mod;
   }
}

void build(int o, int l, int r)
{
   tree[o].l = l, tree[o].r = r;
   if (l == r)
   {
   	tree[o].sum = w[l] % mod;
   	return;
   }

   int mid = (l + r) / 2;
   build(o * 2, l, mid);
   build(o * 2 + 1, mid + 1, r);

   update(o);
}

int ask(int o, int l, int r)
{
   spread(o);

   if (tree[o].l == l && tree[o].r == r)
   	return tree[o].sum;

   int mid = (tree[o].l + tree[o].r) / 2;
   if (r <= mid)
   	return ask(o * 2, l, r);
   else if (l > mid)
   	return ask(o * 2 + 1, l, r);
   else
   	return ask(o * 2, l, mid) + ask(o * 2 + 1, mid + 1, r);
}

void change(int o, int l, int r, int c)
{
   spread(o);
   if (tree[o].l == l && tree[o].r == r)
   {
   	tree[o].tag += c;
   	spread(o);
   	return;
   }

   int mid = (tree[o].l + tree[o].r) / 2;
   if (r <= mid)
   	change(o * 2, l, r, c);
   else if (l > mid)
   	change(o * 2 + 1, l, r, c);
   else
   	change(o * 2, l, mid, c), change(o * 2 + 1, mid + 1, r, c);

   update(o);
}

int ask_range(int u, int v)
{
   int ans = 0;
   while (top[u] != top[v])
   {
   	if (dep[top[v]] < dep[top[u]]) swap(u, v);

   	ans += ask(1, id[top[v]], id[v]);
   	ans %= mod;
   	v = fa[top[v]];
   }

   return ans + ask(1, min(id[u], id[v]), max(id[u], id[v]));
}

int ask_son(int u)
{
   return ask(1, id[u], id[u] + siz[u] - 1) % mod;
}

void change_range(int u, int v, int c)
{
   while (top[u] != top[v])
   {
   	if (dep[top[v]] < dep[top[u]]) swap(u, v);

   	change(1, id[top[v]], id[v], c);
   	v = fa[top[v]];
   }

   change(1, min(id[u], id[v]), max(id[u], id[v]), c);
}

void change_son(int u, int c)
{
   change(1, id[u], id[u] + siz[u] - 1, c);
}

void solve()
{
   scanf("%d%d%d%d", &n, &m, &root, &mod);
   for (int i = 1; i <= n; i++)
   	scanf("%d", &a[i]);

   for (int i = 1; i < n; i++)
   {
   	int u, v;
   	scanf("%d%d", &u, &v);

   	g[u].push_back(v);
   	g[v].push_back(u);
   }

   dfs1(root, root, 1);
   dfs2(root, root);

   build(1, 1, n);

   while (m--)
   {
   	int opt;
   	scanf("%d", &opt);

   	if (opt == 1)
   	{
   		int u, v, c;
   		scanf("%d%d%d", &u, &v, &c);

   		change_range(u, v, c);
   	}
   	else if (opt == 2)
   	{
   		int u, v;
   		scanf("%d%d", &u, &v);

   		printf("%d\n", ask_range(u, v) % mod);
   	}
   	else if (opt == 3)
   	{
   		int u, c;
   		scanf("%d%d", &u, &c);

   		change_son(u, c);
   	}
   	else if (opt == 4)
   	{
   		int u;
   		scanf("%d", &u);

   		printf("%d\n", ask_son(u) % mod);
   	}
   }
}

int main()
{
//	freopen("Testin.txt", "r", stdin);
   //freopen("Testout.txt", "w", stdout);
   solve();
   return 0;
}

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双倍经验

• 无