題意
Sol
直接考慮點分治+hash匹配
設(up[i])表示(dep \% M = i)的從下往上恰好與前(i)位匹配的個數
(down)表示(dep \% M = i)的從上往下恰好與後(i)位匹配的個數
暴力轉移即可
複雜度:(O(nlog^2n)??)
程式碼寫起來有一車邊界
#include<bits/stdc++.h>
#define ull unsigned long long
#define LL long long
#define int long long
#define siz(v) ((int)v.size())
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e10 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < `0` || c > `9`) {if(c == `-`) f = -1; c = getchar();}
while(c >= `0` && c <= `9`) x = x * 10 + c - `0`, c = getchar();
return x * f;
}
int N, M, Root, siz[MAXN], mx[MAXN], Siz, dep[MAXN], up[MAXN], down[MAXN], su[MAXN], sd[MAXN];
LL ans;
bool det[MAXN];
char a[MAXN], b[MAXN];
vector<int> v[MAXN];
ull hs[MAXN], hp[MAXN], po[MAXN], base = 1331;
map<ull, bool> mp;
void FindRoot(int x, int fa) {
siz[x] = 1; mx[x] = 0;
for(int i = 0; i < siz(v[x]); i++) {
int to = v[x][i];
if(to == fa || det[to]) continue;
FindRoot(to, x);
siz[x] += siz[to]; mx[x] = max(mx[x], siz[to]);
}
mx[x] = max(mx[x], Siz - siz[x]);
if(mx[x] < mx[Root]) Root = x;
}
int dfs(int x, int fa, ull now) {
siz[x] = 1;
dep[x] = dep[fa] + 1;
now = now * base + a[x];
if(hp[dep[x]] == now) up[(dep[x] - 1) % M + 1]++, ans += sd[M - (dep[x] - 1) % M];
if(hs[dep[x]] == now) down[(dep[x] - 1) % M + 1]++, ans += su[M - (dep[x] - 1) % M];
// printf("%d %d
", x, ans);
int td =1;
for(int i = 0; i < siz(v[x]); i++) {
int to = v[x][i];
if(to == fa || det[to]) continue;
td = max(td, dfs(to, x, now) + 1);
siz[x] += siz[to];
}
return td;
}
void work(int x) {
int tk = 0, tmp = 0;
det[x] = 1; dep[x] = 1; su[1] = sd[1] = 1;//tag;
for(int i = 0; i < siz(v[x]); i++) {
int to = v[x][i];
if(det[to]) continue;
tk = min(M, dfs(to, x, a[x]) + 1), tmp = max(tmp, tk);
for(int j = 1; j <= tk; j++) su[j] += up[j], sd[j] += down[j], up[j] = down[j] = 0;
}
for(int i = 1; i <= tmp; i++) su[i] = sd[i] = 0;
for(int i = 0; i < siz(v[x]); i++) {
int to = v[x][i];
if(to == x || det[to]) continue;
Siz = siz[to]; Root = 0; FindRoot(to, x);
work(Root);
}
}
void init() {
for(int i = 1; i <= N; i++) v[i].clear();
memset(det, 0, sizeof(det));
memset(siz, 0, sizeof(siz));
memset(mx, 0, sizeof(mx));
ans = 0;
}
void solve() {
N = read(); M = read();
init();
scanf("%s", a + 1);
for(int i = 1; i <= N - 1; i++) {
int x = read(), y = read();
v[x].push_back(y); v[y].push_back(x);
}
for(int i = 1; i <= N; i++) reverse(v[i].begin(), v[i].end());
scanf("%s", b + 1); po[0] = 1;
for(int i = 1; i <= N; i++) {
hp[i] = hp[i - 1] + b[(i - 1) % M + 1] * po[i - 1];
hs[i] = hs[i - 1] + b[M - (i - 1) % M] * po[i - 1];
po[i] = base * po[i - 1];
}
Siz = N; mx[0] = INF; Root = 0; FindRoot(1, 0);
work(1);
printf("%d
", ans);
}
signed main() {
freopen("a.in", "r", stdin);
for(int T = read(); T; T--, solve());
return 0;
}