Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P6329 【模板】点分树 | 震波

P6329 【模板】点分树 | 震波

考虑点分树,点分树的树高为 $O(\log n)$ ,这样对于树上的暴力跳父亲的复杂度是可以接受的。求距离一个点距离为 $k$ 的点的权值和,对于这个点的子树内的点可以用树状数组统计出答案。对于这个点子树外的,考虑暴力跳父亲,每跳一个点后统计外面的答案,即该父亲子树内的答案减去从儿子来的答案。这样,对于一个点需要维护两个树状数组。

查看代码
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#include <cstdio>
#include <vector>
using namespace std;
typedef long long LL;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 1e5 + 10;
bool vis[N];
int n, m, w[N], p[N];
vector<int> g[N];
vector<LL> tr[N][2];
template <class Type>
void chkmin(Type &x, Type k)
{
(x > k) && (x = k);
}
template <class Type>
void chkmax(Type &x, Type k)
{
(x < k) && (x = k);
}
namespace LCA
{
const int K = 20;
int d[N];
int stmp, id[N], st[N << 1][K], lg[N << 1];
void dfs(int x, int fa)
{
st[++stmp][0] = x;
d[x] = d[fa] + 1;
id[x] = stmp;
for (int i : g[x])
if (i ^ fa)
{
dfs(i, x);
st[++stmp][0] = x;
}
}
int dmin(int x, int y)
{
return d[x] < d[y] ? x : y;
}
void init()
{
dfs(1, 0);
for (int i = 2; i <= stmp; i++)
lg[i] = lg[i >> 1] + 1;
for (int k = 1; (1 << k) <= stmp; k++)
for (int i = 1; i + (1 << k) - 1 <= stmp; i++)
st[i][k] = dmin(st[i][k - 1], st[i + (1 << k - 1)][k - 1]);
}
int dis(int a, int b)
{
int x = id[a], y = id[b];
if (x > y)
swap(x, y);
int k = lg[y - x + 1], lca = dmin(st[x][k], st[y - (1 << k) + 1][k]);
return d[a] + d[b] - 2 * d[lca];
}
}
int Size(int x, int fa)
{
if (vis[x])
return 0;
int res = 1;
for (int i : g[x])
i ^ fa && (res += Size(i, x));
return res;
}
int WeightCentre(int x, int fa, int tot, int &wc)
{
if (vis[x])
return 0;
int sum = 1, mx = 0;
for (int i : g[x])
if (i != fa)
{
int t = WeightCentre(i, x, tot, wc);
chkmax(mx, t);
sum += t;
}
chkmax(mx, tot - sum);
mx <= tot / 2 && (wc = x);
return sum;
}
void modify(int x, int k, vector<LL> &tr)
{
for (x++; x < tr.size(); x += x & -x)
tr[x] += k;
}
LL query(int x, vector<LL> &tr)
{
LL res = 0;
for (chkmin(++x, (int)tr.size() - 1); x; x -= x & -x)
res += tr[x];
return res;
}
void calc(int x)
{
if (vis[x])
return;
int sz = Size(x, 0);
WeightCentre(x, -1, sz, x);
vis[x] = true;
tr[x][0].resize(sz + 2), tr[x][1].resize(sz + 2);
for (int i : g[x])
if (!vis[i])
{
int t;
WeightCentre(i, 0, Size(i, -1), t);
p[t] = x;
}
for (int i : g[x])
calc(i);
}
void change(int x, int k)
{
for (int i = x; i; i = p[i])
modify(LCA::dis(x, i), k, tr[i][0]);
for (int i = x; p[i]; i = p[i])
modify(LCA::dis(x, p[i]), k, tr[i][1]);
}
int main()
{
read(n), read(m);
for (int i = 1; i <= n; i++)
read(w[i]);
for (int i = 1, a, b; i < n; i++)
{
read(a), read(b);
g[a].push_back(b);
g[b].push_back(a);
}
LCA::init();
calc(1);
for (int i = 1; i <= n; i++)
change(i, w[i]);
for (int op, last = 0; m; m--)
{
read(op);
if (!op)
{
int x, d;
read(x), read(d);
x ^= last, d ^= last;
LL res = 0;
res += query(d, tr[x][0]);
for (int i = x; p[i]; i = p[i])
{
int k = LCA::dis(x, p[i]);
if (d >= k)
res += query(d - k, tr[p[i]][0]) - query(d - k, tr[i][1]);
}
write(last = res), puts("");
}
else
{
int x, k;
read(x), read(k);
x ^= last, k ^= last;
change(x, k - w[x]);
w[x] = k;
}
}
return 0;
}