【BZOJ 3052】【WC 2013】糖果公园
对树的dfs序分块,打开了新世界的大门233
第一关键字是l所在的块,第二关键字是r所在的块,第三关键字是时间,分完块后暴力莫队即可
#include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; const int N = 100003; void read(int &k) { k = 0; int fh = 1; char c = getchar(); for(; c < '0' || c > '9'; c = getchar()) if (c == '-') fh = -1; for(; c >= '0' && c <= '9'; c = getchar()) k = (k << 1) + (k << 3) + c - '0'; k = k * fh; } bool vis[N]; ll A[N], ans = 0; struct TIME {int x, y, la;} Time[N]; struct edge {int nxt, to;} E[N << 1]; struct node {int l, r, lca, id, tim;} Q[N]; int cal[N], n, m, q, V[N], W[N], f[N][17], deep[N], bel[N << 1]; int point[N], color[N], colorchange[N], pos[N << 1], L[N], R[N], cnt = 0; void ins(int x, int y) {E[++cnt].nxt = point[x]; E[cnt].to = y; point[x] = cnt;} void _(int x, int fa) { pos[L[x] = ++cnt] = x; for(int i = 1; i <= 16; ++i) {f[x][i] = f[f[x][i - 1]][i - 1]; if (f[x][i] == 0) break;} for(int tmp = point[x]; tmp; tmp = E[tmp].nxt) if (E[tmp].to != fa) {deep[E[tmp].to] = deep[x] + 1; f[E[tmp].to][0] = x; _(E[tmp].to, x);} pos[R[x] = ++cnt] = x; } int LCA(int x, int y) { if (deep[x] < deep[y]) swap(x, y); int d = deep[x] - deep[y]; for(int i = 0; i <= 16; ++i) if (d & (1 << i)) x = f[x][i]; if (x == y) return x; for(int i = 16; i >= 0; --i) if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i]; return f[x][0]; } void update(int x) { if (vis[x]) {ans -= 1ll * V[color[x]] * W[cal[color[x]]]; --cal[color[x]];} else {++cal[color[x]]; ans += 1ll * V[color[x]] * W[cal[color[x]]];} vis[x] = !vis[x]; } void change(int a, int b) { if (vis[a]) {update(a); color[a] = b; update(a);} else color[a] = b; } void Timechange(int &a, int b) { while (a < b) {++a; change(Time[a].x, Time[a].y);} while (a > b) {change(Time[a].x, Time[a].la); --a;} } bool cmp(node A, node B) {return bel[A.l] == bel[B.l] ? (bel[A.r] == bel[B.r] ? A.tim < B.tim : A.r < B.r) : A.l < B.l;} int main() { read(n); read(m); read(q); for(int i = 1; i <= m; ++i) read(V[i]); for(int i = 1; i <= n; ++i) read(W[i]); int op, u, v, timecount = 0, tmp = 0, lca; for(int i = 1; i < n; ++i) { read(u); read(v); ins(u, v); ins(v, u); } for(int i = 1; i <= n; ++i) read(color[i]), colorchange[i] = color[i]; cnt = 0; _(1, 0); for(int i = 1; i <= q; ++i) { read(op); read(u); read(v); if (op == 0) {Time[++timecount] = (TIME) {u, v, colorchange[u]}; colorchange[u] = v;} else { if (L[u] > L[v]) swap(u, v); lca = LCA(u, v); if (lca == u) Q[++tmp] = (node) {L[u], L[v], 0, tmp, timecount}; else Q[++tmp] = (node) {R[u], L[v], lca, tmp, timecount}; } } int nn = n << 1, sq = pow(nn, 2.0 / 3) * 0.5, sign = 0; cnt = 1; for(int i = 1; i <= nn; ++i) { bel[i] = sign; ++cnt; if (cnt > sq) {cnt = 1; ++sign;} } sort(Q + 1, Q + tmp + 1, cmp); int l = 1, r = 0, now = 0, tol, tor; for(int i = 1; i <= tmp; ++i) { tol = Q[i].l; tor = Q[i].r; Timechange(now, Q[i].tim); while (l < tol) update(pos[l++]); while (l > tol) update(pos[--l]); while (r < tor) update(pos[++r]); while (r > tor) update(pos[r--]); if (Q[i].lca) update(Q[i].lca); A[Q[i].id] = ans; if (Q[i].lca) update(Q[i].lca); } for(int i = 1; i <= tmp; ++i) printf("%lld\n", A[i]); return 0; }
dfs序分块战术核导弹速度超快~
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