Max Flow

Max Flow

题目描述

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 K pairs of stalls (1≤K≤100,000). For the ith such pair, you are told two stalls si and ti, 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 K 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_i to t_i, then it counts as being pumped through the endpoint stalls s_i and t_i, as well as through every stall along the path between them.

输入

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 s and t describing the endpoint stalls of a path through which milk is being pumped.

输出

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

样例输入

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

样例输出

9分析：Tarjan+差分思想；　　　难点在于怎么处理树上的区间加减问题；　　　把数列差分思想用到树上，区间[i~j]加上k，等价于a[i]+=k,a[j+1]-=k,ans[t]=Σa[p](1=<p<=t)；　　　树上的s，t区间分成两个，设s，t公共祖先为p，一个是s到p的儿子，另一个是t到p，两个区间都差分一下；　　　最后dfs递归处理类似前缀和的答案；代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
const int maxn=1e5+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,k,t,fa[maxn],ans[maxn],vis[maxn],faa[maxn],ma;
vi a[maxn],query[maxn];
int find(int x)
{
    return faa[x]==x?x:faa[x]=find(faa[x]);
}
void dfs(int now,int pre)
{
    for(int x:a[now])
    {
        if(x!=pre)
        {
            dfs(x,now);
            fa[x]=now;
        }
    }
}
void dfs1(int now)
{
    vis[now]=1;
    for(int x:query[now])
    {
        if(vis[x])
        {
            int p=find(x);
            ans[p]--,ans[fa[p]]--;
        }
    }
    for(int x:a[now])
    {
        if(!vis[x])
        {
            dfs1(x);
            faa[x]=now;
        }
    }
}
void dfs2(int now,int pre)
{
    for(int x:a[now])
    {
        if(x!=pre)
        {
            dfs2(x,now);
            ans[now]+=ans[x];
        }
    }
    ma=max(ma,ans[now]);
}
int main()
{
    int i,j;
    scanf("%d%d",&n,&k);
    rep(i,1,n)faa[i]=i;
    rep(i,1,n-1)
    {
        scanf("%d%d",&j,&t);
        a[j].pb(t),a[t].pb(j);
    }
    rep(i,1,k)
    {
        scanf("%d%d",&j,&t);
        ans[j]++,ans[t]++;
        query[j].pb(t);
        query[t].pb(j);
    }
    dfs(1,-1);
    dfs1(1);
    dfs2(1,-1);
    printf("%d\n",ma);
    //system("pause");
    return 0;
}