求和
题目背景:
分析:LCA
当时一看到被吓了一跳,这是要斯特林数还是要拉格朗日差值啊,然后看到k <= 50的时候,脸上就只剩下了一个滑稽······直接用sum[i][j]表示第i个点到跟的路径上所有点的权值的j次方之和,对于一次询问u, v, k,答案就是sum[u][k] + sum[v][k] - sum[lca(u, v)][k] - sum[father[lca(u, v)]][k],直接做LCA就好了,倍增,链剖,ST表,想用什么用什么。而且据说出题人数据造锅了,貌似直接就是随机数据程度,啥都不想写还可以直接暴力(滑稽······)
Source:
/* created by scarlyw */ #include <cstdio> #include <string> #include <algorithm> #include <cstring> #include <iostream> #include <cmath> #include <cctype> #include <vector> #include <set> #include <queue> #include <ctime> #include <bitset> inline char read() { static const int IN_LEN = 1024 * 1024; static char buf[IN_LEN], *s, *t; if (s == t) { t = (s = buf) + fread(buf, 1, IN_LEN, stdin); if (s == t) return -1; } return *s++; } ///* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = read(), iosig = false; !isdigit(c); c = read()) { if (c == -1) return ; if (c == '-') iosig = true; } for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int OUT_LEN = 1024 * 1024; char obuf[OUT_LEN]; char *oh = obuf; inline void write_char(char c) { if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf; *oh++ = c; } template<class T> inline void W(T x) { static int buf[30], cnt; if (x == 0) write_char('0'); else { if (x < 0) write_char('-'), x = -x; for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48; while (cnt) write_char(buf[cnt--]); } } inline void flush() { fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; } /* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = getchar(), iosig = false; !isdigit(c); c = getchar()) if (c == '-') iosig = true; for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int MAXN = 300000 + 10; const int mod = 998244353; int n, m, x, y, k, lca; std::vector<int> edge[MAXN]; int top[MAXN], son[MAXN], size[MAXN], father[MAXN], dep[MAXN]; int sum[MAXN][60]; inline void add_edge(int x, int y) { edge[x].push_back(y), edge[y].push_back(x); } inline void dfs1(int cur, int fa) { dep[cur] = dep[fa] + 1, father[cur] = fa, size[cur] = 1; for (int i = 0, ret = 1; i <= 50; ++i, ret = (long long)ret * (dep[cur] - 1) % mod) { sum[cur][i] = (ret + sum[fa][i]) % mod; } for (int p = 0; p < edge[cur].size(); ++p) { int v = edge[cur][p]; if (v != fa) { dfs1(v, cur), size[cur] += size[v]; if (size[v] > size[son[cur]]) son[cur] = v; } } } inline void dfs2(int cur, int tp) { top[cur] = tp; if (son[cur]) dfs2(son[cur], tp); for (int p = 0; p < edge[cur].size(); ++p) { int v = edge[cur][p]; if (top[v] == 0) dfs2(v, v); } } inline int query_lca(int u, int v) { while (top[u] != top[v]) (dep[top[u]] > dep[top[v]]) ? u = father[top[u]] : v = father[top[v]]; return (dep[u] > dep[v]) ? v : u; } inline void solve() { R(n); for (int i = 1; i < n; ++i) R(x), R(y), add_edge(x, y); R(m), dfs1(1, 0), dfs2(1, 1); while (m--) { R(x), R(y), R(k), lca = query_lca(x, y); W(((sum[x][k] + sum[y][k] - sum[lca][k] - sum[father[lca]][k]) % mod + mod) % mod), write_char('\n'); } } int main() { freopen("sum.in", "r", stdin); freopen("sum.out", "w", stdout); solve(); flush(); return 0; }