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

Posted 2021-03-08 --hpy-7m

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

  1 #define ios ios_base::sync_with_stdio(0); cin.tie(0);
  2 #include <cstdio>//sprintf islower isupper
  3 #include <cstdlib>//malloc  exit strcat itoa system("cls")
  4 #include <iostream>//pair
  5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin);
  6 #include <bitset>
  7 //#include <map>
  8 //#include<unordered_map>
  9 #include <vector>
 10 #include <stack>
 11 #include <set>
 12 #include <string.h>//strstr substr strcat
 13 #include <string>
 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 15 #include <cmath>
 16 #include <deque>
 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 18 #include <vector>//emplace_back
 19 //#include <math.h>
 20 #include <cassert>
 21 #include <iomanip>
 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 25 //******************
 26 clock_t __START,__END;
 27 double __TOTALTIME;
 28 void _MS(){__START=clock();}
 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
 30 //***********************
 31 #define rint register int
 32 #define fo(a,b,c) for(rint a=b;a<=c;++a)
 33 #define fr(a,b,c) for(rint a=b;a>=c;--a)
 34 #define mem(a,b) memset(a,b,sizeof(a))
 35 #define pr printf
 36 #define sc scanf
 37 #define ls rt<<1
 38 #define rs rt<<1|1
 39 typedef pair<int,int> PII;
 40 typedef vector<int> VI;
 41 typedef unsigned long long ull;
 42 typedef long long ll;
 43 typedef double db;
 44 const db E=2.718281828;
 45 const db PI=acos(-1.0);
 46 const ll INF=(1LL<<60);
 47 const int inf=(1<<30);
 48 const db ESP=1e-9;
 49 const int mod=(int)1e9+7;
 50 const int N=(int)1e5+10;
 51 
 52 int n;
 53 ll val[N];
 54 ll _val[N];
 55 ll er[66];
 56 //=========================================================================================
 57 vector<vector<int> >G(N),GG(N);
 58 
 59 bool cmp(int a,int b)
 60 {
 61     return val[a]-_val[a]>val[b]-_val[b];
 62 }
 63 
 64 ll pre(int v,int u)
 65 {
 66     int sz=G[v].size();
 67     for(int i=0;i<sz;++i)
 68     {
 69         int to=G[v][i];
 70         if(to!=u)
 71         {
 72             GG[v].push_back(to);
 73             val[v]+=pre(to,v);
 74         }
 75     }
 76     return val[v];
 77 }
 78 
 79 ll st,max_;
 80 int cnt;
 81 
 82 void cut(int u,int fa)
 83 {
 84     if(val[u]-_val[u]<=st)return;
 85     int sz=G[u].size();
 86     for(int i=0;i<sz;++i)
 87     {
 88         if(val[u]-_val[u]<=st)return;
 89         int to=G[u][i];
 90         if(to!=fa)
 91             cut(to,u),_val[u]+=_val[to];
 92     }
 93     if(val[u]-_val[u]<=st)return;
 94     sort(GG[u].begin(),GG[u].end(),cmp);
 95     sz=GG[u].size();
 96     for(int i=0;i<sz;++i)
 97     {
 98         int to=GG[u][i];
 99         if(val[u]-_val[u]<=st)
100             return;
101         cnt++;
102         _val[u]+=val[to]-_val[to];
103     }
104 }
105 
106 ll get(int k)
107 {
108     st=er[50];
109     for(int i=50;i>=1;--i)
110     {
111         if(st-er[i]<max_)continue;
112         for(int j=0;j<=n;++j)_val[j]=0;
113         cnt=1;//根节点我手动切
114         st-=er[i];
115         cut(1,0);
116         if(cnt>k)
117             st+=er[i];
118     }
119     return st;
120 }
121 int TOT=0;
122 
123 void solve()
124 {
125     pr("Case #%d: ",++TOT);
126     int k;
127     sc("%d%d",&n,&k);
128     for(int i=1;i<=n;++i)G[i].clear(),GG[i].clear();
129     for(int i=1;i<n;++i)
130     {
131         int u,v;
132         sc("%d%d",&u,&v);
133         G[u].push_back(v);
134         G[v].push_back(u);
135     }
136     max_=0;
137     for(int i=1;i<=n;++i)sc("%lld",&val[i]),max_=max(max_,val[i]);
138     pre(1,0);
139     pr("%lld
",get(k));
140 }
141 
142 int main()
143 {
144     er[1]=1;
145     for(int i=2;i<=65;++i)er[i]=er[i-1]*2;//,pr("%d: %lld
",i,er[i]);
146     int T;
147     sc("%d",&T);
148     while(T--)solve();
149     return 0;
150 }
151 
152 /**************************************************************************************/