題目描述

Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), conveniently numbered 1…N. Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of pipes.

FJ is pumping milk between KK pairs of stalls (1≤K≤100,000). For the iith such pair, you are told two stalls s_isi​and t_iti​, endpoints of a path along which milk is being pumped at a unit rate. FJ is concerned that some stalls might end up overwhelmed with all the milk being pumped through them, since a stall can serve as a waypoint along many of the KK paths along which milk is being pumped. Please help him determine the maximum amount of milk being pumped through any stall. If milk is being pumped along a path from si​ to ti​, then it counts as being pumped through the endpoint stalls si​ and

ti​, as well as through every stall along the path between them.

FJ給他的牛棚的N(2≤N≤50,000)個隔間之間安裝了N-1根管道，隔間編號從1到N。所有隔間都被管道連通了。

FJ有K(1≤K≤100,000)條運輸牛奶的路線，第i條路線從隔間si運輸到隔間ti。一條運輸路線會給它的兩個端點處的隔間以及中間途徑的所有隔間帶來一個單位的運輸壓力，你需要計算壓力最大的隔間的壓力是多少。

輸入輸出格式

輸入格式：

The first line of the input contains N and K.

The next N−1 lines each contain two integers x and y (x≠y) describing a pipe

between stalls x and y.

The next K lines each contain two integers ss and tt describing the endpoint

stalls of a path through which milk is being pumped.

輸出格式：

An integer specifying the maximum amount of milk pumped through any stall in the

barn.

輸入輸出樣例

5 10
3 4
1 5
4 2
5 4
5 4
5 4
3 5
4 3
4 3
1 3
3 5
5 4
1 5
3 4

9

/*
     做這道題還是為了做運輸計劃這題 學學樹上差分啥的
     雖然名字叫做最大流 但是用不上網路流的知識啊
     可以說是樹上差分點修改的模板題了（貌似暴力也可，但時間複雜度肯定不優）
     記錄每一個點訪問的次數，維護一個差分陣列f
     設s為起點 t為終點
     則f[s]++ f[t]++ f[lca]-- f[fa[lca]]--
     具體為什麼洛穀日報寫的很詳細了
     https://rpdreamer.blog.luogu.org/ci-fen-and-shu-shang-ci-fen
*/
#include <bits/stdc++.h>

using namespace std;

const int MAXN = 50010;

int deep[MAXN],f[MAXN][25],lg[MAXN],head[MAXN],cnt;
int n,m,s,ans,ppt[MAXN];

struct node{
    int to,pre;
}G[MAXN<<2];

void add(int from,int to){
    G[++cnt].to = to;
    G[cnt].pre = head[from];
    head[from] = cnt;
}

inline int read() {
     int x = 0,m = 1;
     char ch;
     while(ch < '0' || ch > '9')  {if(ch == '-') m = -1;ch = getchar();}
     while(ch >= '0' && ch <= '9'){x = x*10+ch-'0';ch=getchar();}
     return m * x;
 }//讀入優化哦

inline void dfs1(int u)
{
    for(int i = head[u];i;i = G[i].pre)
    {
        int v = G[i].to;
        if(v != f[u][0])
        {
            f[v][0] = u;
            deep[v] = deep[u] + 1;
            dfs1(v);
        }
    }
}//處理LCA

inline int lca(int u,int v)
{
    if(deep[u] < deep[v]) swap(u,v);
    int dis = deep[u] - deep[v];
    for(register int i = 0;i <= lg[n];i++)
    {
        if((1 << i) & dis) u = f[u][i];
    }
    if(u == v) return u;
    for(register int i = lg[deep[u]];i >= 0;i--)
    {
        if(f[u][i] != f[v][i])
        {
            u = f[u][i];v = f[v][i];
        }
    }
    return f[u][0];
}//倍增LCA

inline void init()
{
    for(register int i = 1;i <= n;i++)
    {
        lg[i] = lg[i-1] + (1 << lg[i-1] + 1 == i);
    }
    for(register int j = 1;j <= lg[n];j++)
    {
        for(register int i = 1;i <= n;i++)
        {
            if(f[i][j-1] != -1)
            f[i][j] = f[f[i][j-1]][j-1];
        }
    }
}//初始化

inline void dfs2(int u,int fa){
    for(int i = head[u];i;i = G[i].pre){
        int v = G[i].to;
        if(v == fa) continue;
        dfs2(v,u);
        ppt[u] += ppt[v];
    }
    ans = max(ans,ppt[u]);
}//把差分陣列加回去（字首和？）

int main()
{
    int x,y;
    n = read();m = read();
    for(register int i = 1;i <= n-1;i++)
    {
        x = read();y = read();
        add(x,y);add(y,x);//樹要雙向加邊
    }
    dfs1(1);
    init();
    for(int i = 1;i <= m;i++){
        x = read();y = read();
        int Lca = lca(x,y);
        ppt[x]++,ppt[y]++,ppt[Lca]--,ppt[f[Lca][0]]--;//樹上差分 不多說
    }
    dfs2(1,0);
    printf("%d\n",ans);
    return 0;
}