Luo Gu P3128 LCA + trees difference

Topic links: point I

Subject description:

FJ to his barn N installed between N-1 of conduits (2≤N≤50,000) compartments, the compartments are numbered from 1 to N. All the compartments are in communication pipe.

There FJ K (1≤K≤100,000) milk transport route, a route leading from the i-th compartment to compartment si ti. Pressure a transport route will give its compartment and all the compartments of the middle route of two endpoints to bring pressure on the transportation of a unit, you need to calculate the maximum pressure compartment is.

The meaning of problems

In an unrooted trees, and continue to a point so that s t path (inclusive) plus a pressure unit, after the k-th pressure, the maximum pressure point, how much pressure.

answer

LCA + tree differential
may wish to identify a root node, a pressure dis array record for each node, if the pressure in the u-> v path, it can be decomposed into u-> lca, lca-> v paths each with two pressure.
Because the differential can be done for Operating range:
If there is a set of numbers: 12,345
array dd Difference after: 11,111
to 2-4 adds 1 to 3-5 minus one
need only direct the operation of the differential array, prefix and then again seek to get the answer:
dd [2] ++, dd [+. 1. 4] -; dd [. 3] -, dd [+. 5. 1] ++ (over a range corresponding to no effect);
operating post differential array dd: 1 2 0 1 0
prefix and: 13,344 after the operation for the final answer.
Also use this idea, the difference in the tree, it becomes a tree difference, where each point needs to u-> root, v-> root of u-> v + 1 point each, equivalent to + 1, and u, v common ancestor to the root of each point minus one.
U After the operation, the most recent common ancestor pressure v becomes zero, so the pressure needs to be recorded separately at the recent common ancestor.
dis [u] ++, dis [ v] ++, dis [lca] - = 2;
with the latest separately recorded dis1 pressure common ancestor: dis1 [lca] ++;
last time and the prefix tree, search again DFS, Incidentally, the maximum value also recorded it.
max_ans = max (max_ans, dis [ cur] + dis1 [cur]);

#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false)
#define ll long long
#define maxn 50005
#define maxq 100005

struct node
{
    int value, order;
};
struct edge
{
    int to, next;
};

edge ff[2 * maxn];
int head[2 * maxn], num_edge;
void add_edge(int from, int to)
{
    ff[++num_edge].next = head[from];
    ff[num_edge].to = to;
    head[from] = num_edge;
}

vector<node>qq[maxq];       //离线存储输入
bool vis[maxn];
int yy[maxn], ans[maxq];     //yy为并查集合并数组

int find(int x)
{
    return yy[x] == x ? x : yy[x] = find(yy[x]);
}


void dfs(int from, int to)
{
    vis[to] = true;
    for (int i = head[to]; i; i = ff[i].next)
        if (ff[i].to != from)
            dfs(to, ff[i].to);
    for (int i = 0; i<qq[to].size(); i++)
        if (vis[qq[to][i].value] && ans[qq[to][i].order] == 0)
            ans[qq[to][i].order] = find(qq[to][i].value);
    yy[to] = find(from);
}

vector<pair<int,int> >V;

int dis[maxn];//树上差分，记录前向点（边）的差分流量
int dis1[maxn];
int max_ans=0;
void dfs2(int fa,int cur)
{
    int sum=0;
    for(int i=head[cur]; i; i=ff[i].next)
    {
        int to=ff[i].to;
        if(fa!=to)
        {
            dfs2(cur,to);
            dis[cur]+=dis[to];
        }
    }
    max_ans=max(max_ans,dis[cur]+dis1[cur]);
}

int main()
{
    IOS;
    int n, m;
    cin>>n>>m;

    for (int i = 0; i<n - 1; i++)
    {
        int x, y;
        cin>>x>>y;
        add_edge(x, y);
        add_edge(y, x);
    }

    for (int i = 0; i<m; i++)
    {
        int x, y;
        cin>>x>>y;

        V.push_back(make_pair(x,y));
        qq[x].push_back((node)
        {
            y, i + 1
        });
        qq[y].push_back((node)
        {
            x, i + 1
        });
    }

    for (int i = 0; i <= n; i++)yy[i] = i;

    dfs(1, 1);

    for(int i=0; i<V.size(); i++)
    {
        int id=i+1;
        int x=V[i].first,y=V[i].second;
        int lca=ans[id];
        dis[x]++,dis[y]++,dis[lca]-=2;
        dis1[lca]++;
        //cout<<dis[x]<<","<<dis[y]<<","<<dis[lca]<<"\n";
    }

    /*for(int i=1; i<=n; i++)
        cout<<dis[i]<<" ";
    cout<<"\n";*/
    dfs2(1,1);

    cout<<max_ans;
    return 0;
}

Deformation Title: Point Differential -> differential edge

If not pressurize each for each point, but for all sides from s-> t in pressurized, how to find it? In fact, such a change, but doing it is also simpler, will change the meaning of dis: dis [i] represents the point before i traffic to the edge. Because of this common ancestor absence offset pressure value is 0, no separate recording of pressure common ancestor. Differential, prefixes, and to get the answer.