题目大意:给一棵$n$个节点的树,每个点有一个值$C_i$,每次询问一条路径$x->y$,求$sumlimits_{c}val_c imes sumlimits_{i=1}^{cnt_c}worth_i(cnt_c=sumlimits_{iin(x->y)}[C_i==c])$。带修改
题解:树上带修莫队,在普通的树上莫队上加一维时间即可
卡点:$res$忘记开$long;long$
C++ Code:
#include <cstdio> #include <algorithm> #define maxn 100010 #define N (maxn << 1) #define bl(x) ((x) >> 11) int n, m, l, r, p; long long res; long long ans[maxn]; int C[maxn], num[maxn]; long long W[maxn], V[maxn]; bool vis[maxn]; int Qcnt, Mcnt; struct Query { int l, r, tim, id, lca; bool addlca; inline bool operator < (const Query &rhs) const { return (bl(l) == bl(rhs.l)) ? (bl(r) == bl(rhs.r) ? tim < rhs.tim : r < rhs.r) : l < rhs.l; } } q[maxn]; inline void swap(int &a, int &b) {a ^= b ^= a ^= b;} struct Modity { int pos, C, tim; inline void modify() { if (vis[pos]) { res -= W[num[::C[pos]]--] * V[::C[pos]]; res += W[++num[C]] * V[C]; } swap(::C[pos], C); } } M[maxn]; int date[N], in[maxn], out[maxn], idx; namespace tree { int head[maxn], cnt; struct Edge { int to, nxt; } e[maxn << 1]; void add(int a, int b) { e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt; e[++cnt] = (Edge) {a, head[b]}; head[b] = cnt; } int fa[maxn]; void dfs(int u) { date[in[u] = ++idx] = u; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (v != fa[u]) { fa[v] = u; dfs(v); } } date[out[u] = ++idx] = u; } } namespace tarjan { int head[maxn], cnt; struct QUERY { int v, nxt, id; } Q[maxn << 1]; void add(int a, int b, int c) { Q[++cnt] = (QUERY) {b, head[a], c}; head[a] = cnt; Q[++cnt] = (QUERY) {a, head[b], c}; head[b] = cnt; } int f[maxn]; void init(int n) { for (int i = 1; i <= n; i++) f[i] = i; } int find(int x) {return (x == f[x] ? x : (f[x] = find(f[x])));} void dfs(int u) { for (int i = tree::head[u]; i; i = tree::e[i].nxt) { int v = tree::e[i].to; if (v != tree::fa[u]) { dfs(v); f[v] = find(u); } } for (int i = head[u]; i; i = Q[i].nxt) q[Q[i].id].lca = find(Q[i].v); } } #define ONLINE_JUDGE #include <cctype> namespace R { int x; #ifdef ONLINE_JUDGE char *ch, op[1 << 26]; inline void init() { fread(ch = op, 1, 1 << 26, stdin); } inline int read() { while (isspace(*ch)) ch++; for (x = *ch & 15, ch++; isdigit(*ch); ch++) x = x * 10 + (*ch & 15); return x; } #else char ch; inline int read() { ch = getchar(); while (isspace(ch)) ch = getchar(); for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15); return x; } #endif } int O; int main() { #ifdef ONLINE_JUDGE R::init(); #endif tarjan::init(n = R::read()); m = R::read(), O = R::read(); for (int i = 1; i <= m; i++) V[i] = R::read(); for (int i = 1; i <= n; i++) W[i] = R::read(); for (int i = 1; i < n; i++) tree::add(R::read(), R::read()); tree::dfs(1); for (int i = 1; i <= n; i++) C[i] = R::read(); for (int i = 1; i <= O; i++) { if (R::read()) { q[++Qcnt].tim = i; tarjan::add(q[Qcnt].l = R::read(), q[Qcnt].r = R::read(), q[Qcnt].id = Qcnt); } else M[++Mcnt].pos = R::read(), M[Mcnt].C = R::read(), M[Mcnt].tim = i; } tarjan::dfs(1); for (int i = 1; i <= Qcnt; i++) { int &l = q[i].l, &r = q[i].r; if (in[l] > in[r]) swap(l, r); l = (q[i].addlca = (q[i].lca != l)) ? out[l] : in[l]; r = in[r]; } std::sort(q + 1, q + Qcnt + 1); l = 1, r = 0, p = 0; for (int i = 1; i <= Qcnt; i++) { while (l > q[i].l) (vis[date[--l]] ^= 1) ? (res += W[++num[C[date[l]]]] * V[C[date[l]]]) : (res -= W[num[C[date[l]]]--] * V[C[date[l]]]); while (r < q[i].r) (vis[date[++r]] ^= 1) ? (res += W[++num[C[date[r]]]] * V[C[date[r]]]) : (res -= W[num[C[date[r]]]--] * V[C[date[r]]]); while (l < q[i].l) (vis[date[l]] ^= 1) ? (res += W[++num[C[date[l]]]] * V[C[date[l++]]]) : (res -= W[num[C[date[l]]]--] * V[C[date[l++]]]); while (r > q[i].r) (vis[date[r]] ^= 1) ? (res += W[++num[C[date[r]]]] * V[C[date[r--]]]) : (res -= W[num[C[date[r]]]--] * V[C[date[r--]]]); while (p < Mcnt && M[p + 1].tim < q[i].tim) M[++p].modify(); while (p && M[p].tim > q[i].tim) M[p--].modify(); ans[q[i].id] = res + (q[i].addlca ? W[num[C[q[i].lca]] + 1] * V[C[q[i].lca]] : 0); } for (int i = 1; i <= Qcnt; i++) printf("%lld ", ans[i]); return 0; }