第13届景驰-埃森哲杯广东工业大学ACM程序设计大赛 B题 跳一跳，很简单的（字符串Hash + 树上路径倍增）

题目链接  2018广东工业大学校赛  Problem B

考虑到每条边的权值变化$26$个时刻之后一定会回到原来的状态。

那么预处理出前$26$个时刻每棵树的形态，对每棵树做一遍字符串哈希。

查询的时候找到满足$x$往上爬$k$步和$y$往上爬$k$步之后面对的边的边权不一样的时候的$k$的最小值。

那么比较这条不一样的边的权值就好了。这个过程用倍增实现即可。

时间复杂度$O(nlogn)$

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)    for (int i(a); i <= (b); ++i)
#define dec(i, a, b)    for (int i(a); i >= (b); --i)

typedef unsigned long long LL;

const int N   = 1e5 + 10;
const int A   = 20;
const LL base = 233;

int T;
int n, q;
int a[N], c[N], deep[N], f[N][A], g[26][N];
LL  val[30];
LL  bin[N], h[29][N];
LL  seed = 87378051;
vector <int> v[N];

inline LL rnd(){ return seed = seed * 17 + 23;}

void dfs(int x, int dep){
	deep[x] = dep;

	for (auto u : v[x]){
		int w = a[u];
		rep(i, 0, 25){
			int z = (w + i * c[u]) % 26;
			g[i][u] = z;
			h[i][u] = h[i][x] * base + val[z];
		}
		dfs(u, dep + 1);
	}
}

int main(){

	rep(i, 0, 26) val[i] = rnd() * rnd() * rnd();
	bin[0] = 1;
	rep(i, 1, 1e5 + 1) bin[i] = bin[i - 1] * base;

	scanf("%d", &T);
	while (T--){
		memset(f, 0, sizeof f);
		memset(h, 0, sizeof h);
		memset(g, 0, sizeof g);
		scanf("%d", &n);
		rep(i, 0, n + 1) v[i].clear();
		rep(i, 2, n){
			int x, k;
			char ch[2];
			scanf("%d%s%d", &x, ch, &k);
			a[i] = ch[0] - 'a';
			c[i] = k;
			f[i][0] = x;
			v[x].push_back(i);
		}

		rep(i, 0, 25) g[i][1] = -1;
		dfs(1, 0);

		rep(i, 1, n) rep(j, 0, 18) f[i][j + 1] = f[f[i][j]][j];

		scanf("%d", &q);
		while (q--){
			int x, y, t;
			scanf("%d%d%d", &x, &y, &t);
			t %= 26;

			int dep = min(deep[x], deep[y]);
			int z = 1, m = 0;
			for (; (z << 1) <= dep; z <<= 1) ++m;

			dec(i, m, 0){
				int fx = f[x][i], fy = f[y][i];
				int now = 1 << i;
				LL hx = h[t][x] - h[t][fx] * bin[now];
				LL hy = h[t][y] - h[t][fy] * bin[now];
				if (hx == hy){
					x = fx;
					y = fy;
				}
			}

			if (g[t][x] < g[t][y]) puts("<");
			else if (g[t][x] > g[t][y]) puts(">");
			else puts("=");
		}
	}

	return 0;

}