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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;
}

【JZOJ3347】树的难题

标签：src   des   amp   ast   using   desc   algo   min   str   

原文地址：https://www.cnblogs.com/horizonwd/p/11166964.html