CodeForces 700B Connecting Universities

统计每一条边的贡献，假设$u$是$v$的父节点，$(u,v)$的贡献为：$v$下面大学个数$f[v]$与$2*k-f[v]$的较小值。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar(); x = ;while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x *  + c - ''; c = getchar();  }
}

const int maxn=;
struct Edge { int u,v,nx; }e[*maxn];
int h[maxn],sz;
int n,m,f[maxn],dis[maxn];
LL sum;

void dfs(int x,int fa)
{
    for(int i=h[x];i!=-;i=e[i].nx)
    {
        if(fa==e[i].v) continue;
        dfs(e[i].v,x); f[x]=f[x]+f[e[i].v];
    }
    for(int i=h[x];i!=-;i=e[i].nx)
    {
        if(fa==e[i].v) continue;
        sum=sum+min(f[e[i].v],*m-f[e[i].v]);
    }
}

void add(int u,int v)
{
    e[sz].u=u; e[sz].v=v; e[sz].nx=h[u]; h[u]=sz++;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=;i<=*m;i++) { int x; scanf("%d",&x); f[x]=; }
    memset(h,-,sizeof h);
    for(int i=;i<n;i++)
    {
        int u,v; scanf("%d%d",&u,&v);
        add(u,v); add(v,u);
    }
    dfs(,); printf("%lld\n",sum);
    return ;
}