洛谷P4427 [BJOI2018]求和 树剖+50个前缀和

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洛谷P4427 [BJOI2018]求和

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标签

• 树链剖分

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简明题意

• 给一棵有根树，节点的权值定义为节点的深度。只需要你支持一种查询操作
  1. 查询树上u到v的点权k次方之 (1 <= k <= 50)

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思路

• 果断树剖，去维护区间k次方和，显然要维护50个数据结构，维护不同的次方和。这里我太笨了我开了50棵线段树然后突然想到区间查和前缀和不就行了，然后换成50个前缀和，就A了…

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注意事项

• 前缀和查询的时候，会涉及减法，这样的话取模就会出现问题。所以涉及到减法的地方，先加上模数再取模，就不会出现问题了~~

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总结

• 无

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AC代码

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

const int maxn = 300000 + 10;
const int mod = 998244353;

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

int qsm(int a, int b)
{
   int ans = 1, base = a;
   while (b)
   {
   	if (b & 1)
   		ans = 1ll * ans * base % mod;
   	b >>= 1;
   	base = 1ll * base * base % mod;
   }
   return ans;
}

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

int top[maxn], cnt, id[maxn], w[maxn];
void dfs2(int u, int topf)
{
   id[u] = ++cnt;
   w[cnt] = dep[u];
   top[u] = topf;

   if (son[u])
   {
   	dfs2(son[u], topf);
   	for (auto& v : g[u])
   		if (v != son[u] && v != fa[u])
   			dfs2(v, v);
   }
}

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

struct Pre
{
   int pre[maxn];

   void build(int k)
   {
   	for (int i = 1; i <= n; i++)
   		pre[i] = (pre[i - 1] + qsm(w[i], k)) % mod;
   }

   int ask(int l, int r)
   {
   	return (pre[r] - pre[l - 1] + mod) % mod;
   }
};

Pre tree[51];

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

   	ans += tree[k].ask(id[top[v]], id[v]);
   	ans %= mod;

   	v = fa[top[v]];
   }
   return (ans + tree[k].ask(min(id[u], id[v]), max(id[u], id[v]))) % mod;
}

void solve()
{
   scanf("%d", &n);
   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(1, 1, 0);
   dfs2(1, 1);

   for (int i = 1; i <= 50; i++)
   	tree[i].build(i);

   scanf("%d", &m);
   while (m--)
   {
   	int u, v, k;
   	scanf("%d%d%d", &u, &v, &k);

   	printf("%d\n", ask_path(u, v, k));
   }
}

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

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双倍经验

• 无