題目描述

Farmer John has installed a new system of N-1N?1 pipes to transport milk between the NN stalls in his barn (2 \leq N \leq 50,0002≤N≤50,000), conveniently numbered 1 \ldots N1…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 \leq K \leq 100,0001≤K≤100,000). For the iith such pair, you are told two stalls s_is?i?? and t_it?i??, 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 s_is?i?? to t_it?i??, then it counts as being pumped through the endpoint stalls s_is?i?? and

t_it?i??, 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 NN and KK.

The next N-1N?1 lines each contain two integers xx and yy (x \ne yx≠y) describing a pipe

between stalls xx and yy.

The next KK 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

題目大意：k條路徑所經節點權值+1，求節點最大權值。

題解:樹剖lca+樹上差分

代碼：

#include<iostream>
#include<cstdio>
#define maxn 50002
using namespace std;

int n,m,sumedge,ans;
int head[maxn],dad[maxn],deep[maxn],size[maxn],top[maxn];
int c[maxn];

struct Edge{
    int x,y,nxt;
    Edge(int x=0,int y=0,int nxt=0):
        x(x),y(y),nxt(nxt){}
}edge[maxn<<1];

void add(int x,int y){
    edge[++sumedge]=Edge(x,y,head[x]);
    head[x]=sumedge;
}

void dfs(int x){
    size[x]=1;deep[x]=deep[dad[x]]+1;
    for(int i=head[x];i;i=edge[i].nxt){
        int v=edge[i].y;
        if(v!=dad[x]){
            dad[v]=x;
            dfs(v);
            size[x]+=size[v];
        }
    }
}

void dfs_(int x){
    int s=0;
    if(!top[x])top[x]=x;
    for(int i=head[x];i;i=edge[i].nxt){
        int v=edge[i].y;
        if(v!=dad[x]&&size[v]>size[s])s=v;
    }
    if(s){
        top[s]=top[x];
        dfs_(s);
    }
    for(int i=head[x];i;i=edge[i].nxt){
        int v=edge[i].y;
        if(v!=dad[x]&&v!=s)dfs_(v);
    }
}

int lca(int x,int y){
    for(;top[x]!=top[y];){
        if(deep[top[x]]>deep[top[y]])swap(x,y);
        y=dad[top[y]];
    }
    if(deep[x]>deep[y])swap(x,y);
    return x;
}

void dfs__(int x){
    for(int i=head[x];i;i=edge[i].nxt){
        int v=edge[i].y;
        if(v!=dad[x]){
            dfs__(v);
            c[x]+=c[v];
            ans=max(ans,c[x]);
        }
    }
}

int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<n;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        add(x,y);add(y,x);
    }
    dfs(1);dfs_(1);
    for(int i=1;i<=m;i++){
        int x,y,gg;
        scanf("%d%d",&x,&y);
        gg=lca(x,y);
        c[x]++;c[y]++;c[gg]--;c[dad[gg]]--;
    }
    dfs__(1);
    printf("%d\n",ans);
    return 0;
}

P3128 [USACO15DEC]最大流Max Flow