P3345 [ZJOI2015]幻想乡战略游戏

发表于 | 阅读数

动态维护，加权点重心。

假设当前决策点为$u$,$v$为其子结点。

显然只有一个$v$可能满足。

此时$ans$在$v$的子树里！看到这个想到点分树。

  • 需要在原来的树结构上寻找决策点，而不是点分树的。
  • 进入下一个子树，显然 可以用点分树的子树。

可以知道答案不会转移过多，所以只需啊计算$\sum dis(i,v)\times d_i$

这个时候就是最普通的点分树了，一层一层往上统计，去除重复的。

复杂度$O(n\log^2n)$

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, ll>
#define mk make_pair
const int N = 2e5 + 10;
const int mod = 1e9 + 7;

int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}
vector<pii> g[N];
vector<pii> dg[N];
int dfn[N], st[N][20], tot, dep[N];
int lg2[N];
void dfs1(int x, int fa, int d)
{
    st[++tot][0] = x;
    dep[x] = d;
    dfn[x] = tot;
    for (pii now : g[x])
    {
        int to = now.first;
        if (to == fa)
            continue;
        dfs1(to, x, d + now.second);
        st[++tot][0] = x;
    }
}
int lower(int u, int v)
{
    return (dep[u] < dep[v]) ? u : v;
}
void Lcainit()
{
    dfs1(1, 0, 0);
    for (int i = 2; i <= tot; i++)
        lg2[i] = lg2[i >> 1] + 1;
    for (int l = 1; (1 << l) <= tot; l++)
    {
        for (int i = 1; i + (1 << l) - 1 <= tot; i++)
            st[i][l] = lower(st[i][l - 1], st[i + (1 << (l - 1))][l - 1]);
    }
}
int lca(int u, int v)
{
    u = dfn[u];
    v = dfn[v];
    if (u > v)
        swap(u, v);
    int i = lg2[v - u + 1], w = (1 << i);
    return lower(st[u][i], st[v - w + 1][i]);
}
ll dis(int u, int v)
{
    return dep[u] + dep[v] - 2 * dep[lca(u, v)];
}
int siz[N], f[N], vis[N], fa[N];

int gcnt, rt;
void GetRoot(int x, int fa)
{
    siz[x] = 1;
    f[x] = 0;
    for (pii now : g[x])
    {
        int to = now.first;
        if (to == fa || vis[to])
            continue;
        GetRoot(to, x);
        siz[x] += siz[to];
        f[x] = max(f[x], siz[to]);
    }
    f[x] = max(f[x], gcnt - siz[x]);
    if (f[x] < f[rt])
        rt = x;
}
void dfs2(int x)
{
    vis[x] = 1;
    for (pii now : g[x])
    {
        int to = now.first;
        if (vis[to])
            continue;
        gcnt = siz[to];
        rt = 0;
        GetRoot(to, x);
        dg[x].push_back(mk(to, rt));

        fa[rt] = x;
        dfs2(rt);
    }
}
ll sum[N];
ll sumf[N];
ll sumd[N];

void update(int x, int val)
{

    sum[x] += val;
    for (int i = x; fa[i]; i = fa[i])
    {

        ll len = dis(x, fa[i]);

        sum[fa[i]] += val;
        sumd[fa[i]] += val * len;
        sumf[i] += val * len;
    }
}
ll calc(int x)
{
    ll res = sumd[x];

    for (int i = x; fa[i]; i = fa[i])
    {
        ll len = dis(x, fa[i]);
        res += len * (sum[fa[i]] - sum[i]);

        res += (sumd[fa[i]] - sumf[i]);
    }
    return res;
}
ll query(int x)
{

    ll tmp = calc(x);
    for (pii now : dg[x])
    {
        int to = now.first;
        if (calc(to) < tmp)
            return query(now.second);
    }
    return tmp;
}
int main()
{
    int n = read(), q = read();
    for (int i = 1; i < n; i++)
    {
        int u = read(), v = read(), w = read();
        g[u].push_back(mk(v, w));
        g[v].push_back(mk(u, w));
    }
    Lcainit();
    rt = 0;
    f[0] = 1e9;
    gcnt = n;
    GetRoot(1, 0);
    int R = rt;
    dfs2(rt);

    while (q--)
    {
        int x = read(), y = read();

        update(x, y);

        printf("%lld\n", query(R));
    }
}