codeforces round364 div2E Connecting Universities

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/*    题目描述：给出一棵树，节点数n<=2e5，每条树边的长度为1，在树上有2k个节点，将这2k个节点两两配对，问所有对                之间距离的最大值。*/#pragma warning(disable:4786)#pragma comment(linker, "/STACK:102400000,102400000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stack>#include<queue>#include<map>#include<set>#include<vector>#include<cmath>#include<string>#include<sstream>#include<bitset>#define LL long long#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define mem(a,x) memset(a,x,sizeof(a))#define lson l,m,x<<1#define rson m+1,r,x<<1|1using namespace std;const int INF = 0x3f3f3f3f;const int mod = 1e9 + 7;const double PI = acos(-1.0);const double eps=1e-6;const int maxn = 2e5 + 5;int isu[maxn] , sz[maxn] , dis[maxn] , kk;LL ans;vector<int>G[maxn];void dfs1(int root , int fa , LL depth){    int siz = G[root].size() , nt;    dis[root] = depth;    if(isu[root]){        sz[root] = 1;        ans += depth;    }    for(int i = 0 ; i<siz ; i++){        nt = G[root][i];        if(nt == fa)    continue;        dfs1(nt , root , depth + 1LL);        sz[root] += sz[nt];    }}void solve(int root , int fa){    if(kk <= 0)        return ;    priority_queue< pair<int , int > >Q;    int siz = G[root].size() , nt;    for(int i = 0 ; i<siz ; i++){        nt = G[root][i];        if(nt == fa)    continue;        if(sz[nt] > 0){            Q.push( make_pair(sz[nt] , nt) );        }    }    if(isu[root]){        Q.push( make_pair(1 , root) );    }    pair<int , int> f1 = Q.top() , f2;    int fir = f1.first , sec = f1.second , sum = 0;    if(Q.size() == 1){        solve(sec , root);    }    else{        Q.pop();        while(!Q.empty()){            f2 = Q.top();   Q.pop();            sum += f2.first;        }        sum = min(sum , kk);        kk -= sum;        ans -= 2LL * dis[root] * sum;        if(kk > 0)            solve(sec , root);    }}int main(){    int n , k , x , y;    scanf("%d %d", &n , &k);    kk = k;    for(int i = 1 ; i <= 2* k ; i++){        scanf("%d", &x);        isu[x] = 1;    }    for(int i = 1 ; i< n ; i++){        scanf("%d %d", &x , &y);        G[x].push_back(y);        G[y].push_back(x);    }    ans = 0;    dfs1(1 , 0 , 0);    solve(1 , 0);    printf("%lld\n" , ans);    return 0;}
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