The 2019 ICPC Asia Shanghai Regional Contest---H Tree Partition 二分答案，从下往上切最大子树

题意：https://ac.nowcoder.com/acm/contest/4370

将一棵树切k-1刀分成k棵树，问这k棵树里最大的权值的最小值是多少

思路：https://www.cnblogs.com/ucprer/p/11931263.html

二分最大值。

dfs割子树，每次从下往上切一个最大的树。

单个dfs表示保证当前节点可切，然后子树dfs完了，sort子树，选最大的切，直到当前u节点也可切

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr strcat
#include <string>
#include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
#include <cassert>
#include <iomanip>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
clock_t __START,__END;
double __TOTALTIME;
void _MS(){__START=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
const db E=2.718281828;
const db PI=acos(-1.0);
const ll INF=(1LL<<60);
const int inf=(1<<30);
const db ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e5+10;

int n;
ll val[N];
ll _val[N];
ll er[66];
//=========================================================================================
vector<vector<int> >G(N),GG(N);

bool cmp(int a,int b)
{
    return val[a]-_val[a]>val[b]-_val[b];
}

ll pre(int v,int u)
{
    int sz=G[v].size();
    for(int i=0;i<sz;++i)
    {
        int to=G[v][i];
        if(to!=u)
        {
            GG[v].push_back(to);
            val[v]+=pre(to,v);
        }
    }
    return val[v];
}

ll st,max_;
int cnt;

void cut(int u,int fa)
{
    if(val[u]-_val[u]<=st)return;
    int sz=G[u].size();
    for(int i=0;i<sz;++i)
    {
        if(val[u]-_val[u]<=st)return;
        int to=G[u][i];
        if(to!=fa)
            cut(to,u),_val[u]+=_val[to];
    }
    if(val[u]-_val[u]<=st)return;
    sort(GG[u].begin(),GG[u].end(),cmp);
    sz=GG[u].size();
    for(int i=0;i<sz;++i)
    {
        int to=GG[u][i];
        if(val[u]-_val[u]<=st)
            return;
        cnt++;
        _val[u]+=val[to]-_val[to];
    }
}

ll get(int k)
{
    st=er[50];
    for(int i=50;i>=1;--i)
    {
        if(st-er[i]<max_)continue;
        for(int j=0;j<=n;++j)_val[j]=0;
        cnt=1;//根节点我手动切
        st-=er[i];
        cut(1,0);
        if(cnt>k)
            st+=er[i];
    }
    return st;
}
int TOT=0;

void solve()
{
    pr("Case #%d: ",++TOT);
    int k;
    sc("%d%d",&n,&k);
    for(int i=1;i<=n;++i)G[i].clear(),GG[i].clear();
    for(int i=1;i<n;++i)
    {
        int u,v;
        sc("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    max_=0;
    for(int i=1;i<=n;++i)sc("%lld",&val[i]),max_=max(max_,val[i]);
    pre(1,0);
    pr("%lld
",get(k));
}

int main()
{
    er[1]=1;
    for(int i=2;i<=65;++i)er[i]=er[i-1]*2;//,pr("%d: %lld
",i,er[i]);
    int T;
    sc("%d",&T);
    while(T--)solve();
    return 0;
}

/**************************************************************************************/