[BJOI2018]求和

Posted 2021-01-15 mrclr

嘟嘟嘟

 

首先看到k只有50，那么就可以开一个数组预处理出来。

sum[u][k]表示节点u到根节点所有节点深度的k次方和，dfs一遍就都搞出来了，预处理复杂度O(n * 50)（快速幂复杂度不计了）。

查询就是lca复杂度，对于路径(x, y)，令z = lca(x, y)，则ans(x, y)  = sum[x][k] + sum[y][k] - sum[z][k] - sum[fa[z]][k]。注意取模别出负数。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cmath>
  4 #include<algorithm>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<cctype>
  8 #include<vector>
  9 #include<stack>
 10 #include<queue>
 11 using namespace std;
 12 #define enter puts("") 
 13 #define space putchar(‘ ‘)
 14 #define Mem(a, x) memset(a, x, sizeof(a))
 15 #define rg register
 16 typedef long long ll;
 17 typedef double db;
 18 const int INF = 0x3f3f3f3f;
 19 const db eps = 1e-8;
 20 const int maxn = 3e5 + 5;
 21 const ll mod = 998244353;
 22 inline ll read()
 23 {
 24       ll ans = 0;
 25     char ch = getchar(), last = ‘ ‘;
 26     while(!isdigit(ch)) {last = ch; ch = getchar();}
 27     while(isdigit(ch)) {ans = ans * 10 + ch - ‘0‘; ch = getchar();}
 28     if(last == ‘-‘) ans = -ans;
 29     return ans;
 30 }
 31 inline void write(ll x)
 32 {
 33     if(x < 0) x = -x, putchar(‘-‘);
 34     if(x >= 10) write(x / 10);
 35     putchar(x % 10 + ‘0‘);
 36 }
 37 
 38 int n, m;
 39 struct Edge
 40 {
 41     int nxt, to;
 42 }e[maxn << 1];
 43 int head[maxn], ecnt = -1;
 44 void addEdge(int x, int y)
 45 {
 46     e[++ecnt] = (Edge){head[x], y};
 47     head[x] = ecnt;
 48 }
 49 
 50 ll quickpow(ll a, int b)
 51 {
 52     a %= mod;
 53     ll ret = 1;
 54     for(; b; a = a * a % mod, b >>= 1) 
 55         if(b & 1) ret = ret * a % mod;
 56     return ret;
 57 }
 58 
 59 int dis[maxn];
 60 ll sum[maxn][55];
 61 void dfs(int now, int f)
 62 {
 63     for(int i = head[now]; i != -1; i = e[i].nxt)
 64     {
 65         if(e[i].to == f) continue;
 66         dis[e[i].to] = dis[now] + 1;
 67         for(int j = 1; j <= 50; ++j) sum[e[i].to][j] = (sum[now][j] + quickpow(dis[e[i].to], j)) % mod;
 68         dfs(e[i].to, now);
 69     }
 70 }
 71 
 72 const int N = 22;
 73 int fa[maxn][25];
 74 void dfs2(int now, int f)
 75 {
 76     for(int i = 1; i <= N; ++i)
 77         fa[now][i] = fa[fa[now][i - 1]][i - 1];
 78     for(int i = head[now]; i != -1; i = e[i].nxt)
 79     {
 80         if(e[i].to == f) continue;
 81         fa[e[i].to][0] = now;
 82         dfs2(e[i].to, now);
 83     }
 84 }
 85 int lca(int x, int y)
 86 {
 87     if(dis[x] < dis[y]) swap(x, y);
 88     for(int i = N; i >= 0; --i)
 89         if(dis[x] - (1 << i) >= dis[y]) x = fa[x][i];
 90     if(x == y) return x;
 91     for(int i = N; i >= 0; --i)
 92         if(fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
 93     return fa[x][0];
 94 }
 95 
 96 ll solve(int x, int y, int k)
 97 {
 98     int z = lca(x, y);
 99     return (sum[x][k] + sum[y][k] - (sum[z][k] + sum[fa[z][0]][k]) % mod + mod) % mod;
100 }
101 
102 int main()
103 {
104     Mem(head, -1);
105     n = read();
106     for(int i = 1; i < n; ++i)
107     {
108         int x = read(), y = read();
109         addEdge(x, y); addEdge(y, x);
110     }
111       dfs(1, 0);
112       dfs2(1, 0);
113       m = read();
114       for(int i = 1; i <= m; ++i)
115       {
116           int x = read(), y = read(), k = read();
117           write(solve(x, y, k)), enter;
118       }
119       return 0;
120 }

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