（淀粉质）P2634 [国家集训队]聪聪可可 and P3806 多次离线查询树上距离为k的点对是否存在

一句话点分治：每次找到树的重心，暴力或者不暴力处理以这个重心为分界点的 各个子树 之间 产生的贡献，然后删除这个点，产生很多树，对每棵子树重复上述操作。。。

有点像dsu on tree，都是通过均摊来实现nlogn

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P3806

提供一种离线处理询问的做法：

点分的过程中在对root的单独处理时，每处理一个子树，就用由root其他子树构成的桶去离线处理询问是否存在。最后处理完root的所有子树相互之间产生的贡献后，把桶清空。

void cal(int from)
{
    tmp.clear();
    int last, now;
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to, cost = e[i].cost;
        if (vis[to])continue;
        last = tmp.size();
        //getdis是把子树中得到的路径长度加入tmp中，并得到tmp中的起点last和终点now（暂时不clear，留着最后清空桶）
        getdis(to, from, cost);
        now = tmp.size() - 1;
        //makeans
        for (int i = 1; i <= m; i++)
        {
            int ans = q[i];//当前询问的值
            if (ok[i])continue;//已存在，跳过
            for (int j = last; j <= now; j++)//遍历当前的子树可以得到的路径长度
                if (ans >= tmp[j] && cnt[ans - tmp[j]])//桶中有这样的值
                {
                    ok[i] = 1;//存在
                    break;
                }
        }
        //insert_cnt
        for (int i = last; i <= now; i++)
            if (tmp[i] <= 10000000)
                cnt[tmp[i]] = 1;
    }
    //clear_cnt
    for (auto i : tmp)
        if (i <= 10000000)
            cnt[i] = 0;
}

求重心：

void find(int from, int fa)//找重心
{
    siz[from] = 1, mx[from] = 0;
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to;
        if (to == fa || vis[to])continue;
        find(to, from);
        siz[from] += siz[to];
        mx[from] = max(mx[from], siz[to]);
    }
    mx[from] = max(mx[from], tot - siz[from]);
    if (mx[from] < mx[root])
        root = from;
}

点分：

void divid(int from)
{
    vis[from] = 1;
    cal(from);
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to;
        if (vis[to])continue;
        tot = siz[to];
        root = 0;
        find(to, from);
        divid(root);
    }
}

最终代码：

#include<bits/stdc++.h>
#define ll long long
#define fastio ios::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
using namespace std;

const ll mod = 998244353;
const int maxn = 1e5 + 5;
const int inf = 1e9 + 7;
const ll lnf = 1e18;

struct edge {
    int to, cost, next;
}e[maxn << 1];

int edge_cnt = 0, head[maxn];
void add(int from, int to,int cost)
{
    e[++edge_cnt] = { to,cost, head[from] };
    head[from] = edge_cnt;
}

int siz[maxn], tot, mx[maxn];
int root = 0;
bool vis[maxn];

int n, m;

void find(int from, int fa)//找重心
{
    siz[from] = 1, mx[from] = 0;
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to;
        if (to == fa || vis[to])continue;
        find(to, from);
        siz[from] += siz[to];
        mx[from] = max(mx[from], siz[to]);
    }
    mx[from] = max(mx[from], tot - siz[from]);
    if (mx[from] < mx[root])
        root = from;
}

int q[105], ok[105];

bool cnt[10000005];
vector<int>tmp;

void getdis(int from, int fa, int res)
{
    tmp.push_back(res);
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to, cost = e[i].cost;
        if (vis[to] || to == fa)continue;
        getdis(to, from, res + cost);
    }
}

void cal(int from)
{
    tmp.clear();
    int last, now;
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to, cost = e[i].cost;
        if (vis[to])continue;
        last = tmp.size();
        getdis(to, from, cost);
        /*for (auto i : tmp)
            cout << i << " ";
        cout << endl;*/
        now = tmp.size() - 1;

        //makeans
        for (int i = 1; i <= m; i++)
        {
            int ans = q[i];//当前询问的值
            if (ok[i])continue;//已存在，跳过
            for (int j = last; j <= now; j++)//遍历当前的子树可以得到的路径长度
                if (ans >= tmp[j] && cnt[ans - tmp[j]])//桶中有这样的值
                {
                    ok[i] = 1;//存在
                    break;
                }
        }
        //insert_cnt
        for (int i = last; i <= now; i++)
            if (tmp[i] <= 10000000)
                cnt[tmp[i]] = 1;
    }
    //clear_cnt
    for (auto i : tmp)
        if (i <= 10000000)
            cnt[i] = 0;
}

void divid(int from)
{
    vis[from] = 1;
    cal(from);
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to;
        if (vis[to])continue;
        tot = siz[to];
        root = 0;
        find(to, from);
        divid(root);
    }
}

int main()
{
    fastio;
    memset(head, -1, sizeof(head));
    cin >> n >> m;
    for (int i = 1; i < n; i++)
    {
        int x, y, z;
        cin >> x >> y >> z;
        add(x, y, z);
        add(y, x, z);
    }
    for (int i = 1; i <= m; i++)
        cin >> q[i];
    mx[0] = n;
    cnt[0] = 1;
    root = 0;
    find(1, 0);
    divid(root);
    for (int i = 1; i <= m; i++)
        if (ok[i])
            cout << "AYE" << endl;
        else
            cout << "NAY" << endl;
    return 0;

}

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P2634 [国家集训队]聪聪可可

模版题，没有任何其他操作，直接开个cnt[3]来点分算贡献就行了。

需要注意一点，除了root子树之间的贡献之外，还有对于root子树上的某个点x，x到root，x和x的贡献。

#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define fastio ios::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
using namespace std;

const ll mod = 998244353;
const int maxn = 1e5 + 5;
const int inf = 1e9 + 7;
const ll lnf = 1e18;

struct edge {
    int to, cost, next;
}e[maxn << 1];

int edge_cnt = 0, head[maxn];
void add(int from, int to,int cost)
{
    e[++edge_cnt] = { to,cost, head[from] };
    head[from] = edge_cnt;
}

int siz[maxn], tot, mx[maxn];
int root = 0;
bool vis[maxn];

int n, m;

void find(int from, int fa)//找重心
{
    siz[from] = 1, mx[from] = 0;
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to;
        if (to == fa || vis[to])continue;
        find(to, from);
        siz[from] += siz[to];
        mx[from] = max(mx[from], siz[to]);
    }
    mx[from] = max(mx[from], tot - siz[from]);
    if (mx[from] < mx[root])
        root = from;
}

int cnt[4];
vector<int>tmp;

void getdis(int from, int fa, int res)
{
    tmp.push_back(res % 3);
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to, cost = e[i].cost;
        if (vis[to] || to == fa)continue;
        getdis(to, from, res + cost);
    }
}

int res = 0;

void cal(int from)
{
    tmp.clear();
    int last, now;
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to, cost = e[i].cost;
        if (vis[to])continue;
        last = tmp.size();
        getdis(to, from, cost);
        now = tmp.size() - 1;
        //makeans
        for (int i = last; i <= now; i++)
        {
            if (tmp[i] == 1)
                res += 2 * cnt[2];
            else if (tmp[i] == 2)
                res += 2 * cnt[1];
            else
                res += 2 * cnt[0] + 2;//加2是root和x以及x和root
        }
        //insert_cnt
        for (int i = last; i <= now; i++)
            cnt[tmp[i]]++;
    }
    //clear_cnt
    cnt[0] = cnt[1] = cnt[2] = 0;
}

void divid(int from)
{
    vis[from] = 1;
    cal(from);
    for (int i = head[from]; ~i; i = e[i].next)
    {
        int to = e[i].to;
        if (vis[to])continue;
        tot = siz[to];
        root = 0;
        find(to, from);
        divid(root);
    }
}

int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}

int main()
{
    fastio;
    memset(head, -1, sizeof(head));
    cin >> n;
    int ans = n * n;
    for (int i = 1; i < n; i++)
    {
        int x, y, z;
        cin >> x >> y >> z;
        add(x, y, z);
        add(y, x, z);
    }
    mx[0] = n;
    root = 0;
    find(1, 0);
    divid(root);
    //cout << res<<" "<<ans << endl;
    res += n;
    int g = gcd(res, ans);
    cout << res / g << "/" << ans / g;
    return 0;

}