hdu 4313 Matrix 树形dp

dpn[x]表示以x为根的子树间互相不联通但是可能和x联通所需要删除的边权
dpy[x]表示以x为根的子树间互相不联通而且不能和x联通所需要删除的边权

遍历每个点，通过它的子节点信息更新它的信息，

当x的子节点y上有机器时: dpn[x]+=dpn[y]+cost;

设临时变量 temp>?=cost;

当子节点y上没有机器时：dpn[x]+=min(dpy[to]+cost,dpn[to]);

临时变量: temp>?=min(dpy[to]+cost,dpn[to])-dpy[to];

最后更新dpy[x];
if(x上没机器) dpy[x]=dpn[x]-temp;
复杂度O(n) 比较简单的中档题

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;

#ifdef _WIN32
typedef __int64 i64;
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
typedef long long i64;
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FFF(i,a)        for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int maxn = 100111;

struct zz
{
	int from;
	int to;
	int cost;
}zx;
int n,k;
vector<zz>g[maxn];
bool you[maxn];
i64 dpn[maxn];   //和x不通 
i64 dpy[maxn];   //和x通
int f[maxn];  

void df1(int now)
{
	int to;
	for(int i=0;i<g[now].size();i++)
	{
		to = g[now][i].to;
		if(f[to]==-1)
		{
			f[to]=now;
			df1(to);
		}	
	}
	return ;
}

void df2(int now)
{
	int to;
	int cost;
	i64 temp = 0;
	for(int i=0;i<g[now].size();i++)
	{
		to = g[now][i].to;
		cost = g[now][i].cost;
		if(to!=f[now])
		{
			
			df2(to);
			if(you[to])
			{
				dpn[now]+=dpn[to]+cost;
				if(cost>temp)
				{
					temp = cost;
				}
			}
			else
			{
				dpn[now]+=min(dpy[to]+cost,dpn[to]);
				if(min(dpy[to]+cost,dpn[to])-dpy[to]>temp)
				{
					temp = min(dpy[to]+cost,dpn[to])-dpy[to];
				}
			}
		}		
	}
	if(!you[now])
	{
		dpy[now]=dpn[now]-temp;
	}	
	return ;
}

i64 start()
{
	for(int i=0;i<n;i++)
	{
		f[i]=-1;
		dpn[i]=0;
		dpy[i]=0;
	}
	f[0]=-2;
	df1(0);
	df2(0);
	if(!you[0])
	{
		return dpy[0];
	}
	else
	{
		return dpn[0];
	}
}	

int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		cin>>n>>k;	
		for(int i=0;i<n;i++)	
		{
			you[i]=false;
			g[i].clear();
		}
		for(int i=1;i<=n-1;i++)
		{
			cin>>zx.from>>zx.to>>zx.cost;
			g[zx.from].pb(zx);
			swap(zx.from,zx.to);
			g[zx.from].pb(zx);
		}	
		int now;
		for(int i=1;i<=k;i++)
		{
			cin>>now;
			you[now]=true;
		}
		cout<<start()<<endl;
	}
    return 0;
}