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bzoj3730: 震波
分析
题目大意：待修改询问树上距离某个点不超过K的所有点的点权和。
这种类似不超过多少距离和，树上所有路径之类的问题一般都是点分治，因为有修改所以是动态的。
对于每个分治重心开一颗以深度为键值的动态开点线段树维护点权和。
每次在分治树上跳即可。
注意因为需要删除子树影响，所以每个节点要额外维护一颗进入节点的子树的线段树。
复杂度
代码
#include<bits/stdc++.h>
const int N = 1e5 + 10;
int ri() {
char c = getchar(); int x = 0, f = 1; for(;c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
for(;c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) - '0' + c; return x * f;
}
int tp, n, sums, r1[N], r2[N], v[N];
int G, mn, pr[N], to[N << 1], nx[N << 1], sz[N], de[N], fa[N];
bool vis[N];
void add(int u, int v) {to[++tp] = v; nx[tp] = pr[u]; pr[u] = tp;}
void adds(int u, int v) {add(u, v); add(v, u);}
struct {
int ls[N * 100], rs[N * 100], sum[N * 100], sz, val;
void Modify(int &p, int L, int R, int x) {
if(!p) p = ++sz; sum[p] += val;
if(L == R) return ; int m = L + R >> 1;
x <= m ? Modify(ls[p], L, m, x) : Modify(rs[p], m + 1, R, x);
}
int Query(int p, int L, int R, int k) {
if(!p || k < L || k > R) return 0;
if(R == k) return sum[p]; int m = L + R >> 1;
return k <= m ? Query(ls[p], L, m, k) : sum[ls[p]] + Query(rs[p], m + 1, R, k);
}
}seg;
struct {
int bin[21], de[N], mn[21][N << 1], in[N], Lg[N << 1], tot;
void Dfs(int u, int fa) {
de[u] = de[fa] + 1; mn[0][++tot] = u; in[u] = tot;
for(int i = pr[u]; i; i = nx[i])
if(to[i] != fa)
Dfs(to[i], u),
mn[0][++tot] = u;
}
int Be(int x, int y) {return de[x] < de[y] ? x : y;}
void Work() {
Dfs(1, 0);
bin[0] = 1; for(int i = 1;i <= 20; ++i) bin[i] = bin[i - 1] << 1;
Lg[0] = -1; for(int i = 1;i <= tot; ++i) Lg[i] = Lg[i >> 1] + 1;
for(int i = 1;bin[i] <= tot; ++i)
for(int j = 1;j + bin[i] - 1 <= tot; ++j)
mn[i][j] = Be(mn[i - 1][j], mn[i - 1][j + bin[i - 1]]);
}
int Lca(int u, int v) {
u = in[u]; v = in[v];
if(u > v) std::swap(u, v);
int t = Lg[v - u + 1];
return Be(mn[t][u], mn[t][v - bin[t] + 1]);
}
int Dist(int u, int v) {
return de[u] + de[v] - (de[Lca(u, v)] << 1);
}
}lca;
void Get(int u, int fa, int &R) {
de[u] = de[fa] + 1; seg.val = v[u];
seg.Modify(R, 0, n - 1, de[u]);
for(int i = pr[u]; i; i = nx[i])
if(to[i] != fa && !vis[to[i]])
Get(to[i], u, R);
}
void Rt(int u, int fa) {
sz[u] = 1; int tp = 0;
for(int i = pr[u]; i; i = nx[i])
if(to[i] != fa && !vis[to[i]]) {
Rt(to[i], u);
sz[u] += sz[to[i]];
tp = std::max(tp, sz[to[i]]);
}
tp = std::max(tp, sums - sz[u]);
if(mn > tp) mn = tp, G = u;
}
void Divide(int u, int in, int ps) {
Get(u, 0, r1[u]);
if(in) Get(in, 0, r2[u]);
vis[u] = true;
for(int i = pr[u]; i; i = nx[i])
if(!vis[to[i]]) {
sums = sz[to[i]] > sz[u] ? ps - sz[u] : sz[to[i]];
G = 0; mn = 1e9; Rt(to[i], 0); fa[G] = u;
Divide(G, to[i], sums);
}
}
int Query(int u, int k) {
int Ans = seg.Query(r1[u], 0, n - 1, k);
for(int x = u; fa[x]; x = fa[x]) {
int Dis = lca.Dist(u, fa[x]);
Ans += seg.Query(r1[fa[x]], 0, n - 1, k - Dis);
Ans -= seg.Query(r2[x], 0, n - 1, k - Dis - 1);
}
return Ans;
}
void Modify(int u, int val) {
seg.val = val - v[u];
seg.Modify(r1[u], 0, n - 1, 0);
for(int x = u;x; x = fa[x]) {
int Dis = lca.Dist(u, fa[x]);
seg.Modify(r1[fa[x]], 0, n - 1, Dis);
seg.Modify(r2[x], 0, n - 1, Dis - 1);
}
v[u] = val;
}
int main() {
de[0] = -1; n = ri(); int m = ri();
for(int i = 1;i <= n; ++i)
v[i] = ri();
for(int i = 1;i < n; ++i)
adds(ri(), ri());
lca.Work();
G = 0; mn = 1e9; sums = n; Rt(1, 0);
Divide(G, 0, sums);
for(int La = 0;m--; ) {
int op = ri(), x = ri() ^ La, y = ri() ^ La;
if(!op) printf("%d\n", La = Query(x, y));
else Modify(x, y);
}
return 0;
}