【JZOJ3347】树的难题

description

---

analysis

• 比较麻烦树形\(DP\)

• 不过这个我还是不算很懂……

• 下次要注意思考，不要怕麻烦

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define MAXN 300005
#define MAXM MAXN*2
#define INF 100000000000007
#define ll long long
#define fo(i,a,b) for (ll i=a;i<=b;++i)
#define fd(i,a,b) for (ll i=a;i>=b;--i)
#define rep(i,a) for (ll i=last[a];i;i=next[i])

using namespace std;

ll last[MAXM],next[MAXM],tov[MAXM],len[MAXM];
ll fa[MAXN],color[MAXN],f[MAXN],g[MAXN],h[MAXN],size[MAXN];
bool bz[MAXN];
ll n,T,tot,pos;
deque<ll>q;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline void link(ll x,ll y,ll z){next[++tot]=last[x],last[x]=tot,tov[tot]=y,len[tot]=z;}
int main()
{
	freopen("T1.in","r",stdin);
	T=read();
	while (T--)
	{
		memset(f,0,sizeof(f));
		memset(g,0,sizeof(g));
		memset(h,0,sizeof(h));
		memset(bz,1,sizeof(bz));
		memset(fa,0,sizeof(fa));
		memset(size,0,sizeof(size));
		memset(last,0,sizeof(last));
		memset(next,0,sizeof(next));
		memset(tov,0,sizeof(tov));
		memset(len,0,sizeof(len));
		tot=pos=0,n=read();
		fo(i,1,n)color[i]=read();
		fo(i,1,n-1)
		{
			ll x=read(),y=read(),z=read();
			link(x,y,z),link(y,x,z);
		}
		q.push_back(1),bz[1]=0;
		while (q.size()<n)
		{
			ll x=q[pos++];
			rep(i,x)if (bz[tov[i]])++size[x],fa[tov[i]]=x,bz[tov[i]]=0,q.push_back(tov[i]);
		}
		while (q.size())
		{
			ll x=q[q.size()-1];q.pop_back();
			if (color[x])//白灰
			{
				f[x]=0;
				rep(i,x)if (fa[tov[i]]==x)
				{
					ll tmp=min(g[tov[i]],h[tov[i]])+len[i];
					f[x]+=min(f[tov[i]],tmp);
				}
			}
			else f[x]=INF;
			if (color[x]^1)//黑灰
			{
				g[x]=0;
				rep(i,x)if (fa[tov[i]]==x)
				{
					ll tmp=min(f[tov[i]],h[tov[i]])+len[i];
					g[x]+=min(g[tov[i]],tmp);
				}
				h[x]=INF;
				rep(i,x)if (fa[tov[i]]==x)
				{
					ll tmp=min(f[tov[i]],h[tov[i]])+len[i];
					tmp=h[tov[i]]+g[x]-min(g[tov[i]],tmp);
					h[x]=min(h[x],tmp);
				}
			}
			else//白
			{
				g[x]=INF,h[x]=0;
				rep(i,x)if (fa[tov[i]]==x)
				{
					ll tmp=min(f[tov[i]],h[tov[i]])+len[i];
					h[x]+=min(g[tov[i]],tmp);
				}
			}
		}
		printf("%lld\n",min(f[1],min(g[1],h[1])));
	}
	return 0;
}