旅行-树形DP

Posted 2020-10-02 dancer16

60分：对于每个点为根的情况做一遍DP，s,g表示从小C，A从这个点走出去的路程

100分，树形DP对于每个点的路径可以由其父亲转来，这样的DP就需要比较好的技巧了。

以前有道差不多的叫星座的题目来着。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define ll long long 
int n,x,y,z,num;
int head[300005],next[600005],to[600005],val[600005];
ll f[300005],g[300005],fs[300005],gs[300005],ans[300005]; 
void add_edge(int u,int v,int c)
{
    to[++num]=v;
    val[num]=c;
    next[num]=head[u];
    head[u]=num; 
}
void dp(int u,int fa)
{
    bool son=0;
    for(int edge=head[u];edge;edge=next[edge])
    if(to[edge]!=fa)
    {
        son=1;
        dp(to[edge],u);
        f[u]=max(f[u],g[to[edge]]+val[edge]);
        g[u]=min(g[u],f[to[edge]]+val[edge]);
    }
    if(!son) g[u]=0;
 }
void DP(int u,int fa)
{
    ll s1=1e16,s2=1e16;
    int p1=0,p2=0;
    for(int edge=head[u];edge;edge=next[edge])
    if(to[edge]!=fa)
    {
        if(f[to[edge]]+val[edge]<s1)
        {
            s2=s1;
            s1=f[to[edge]]+val[edge];
            p2=p1;
            p1=to[edge];
        }else
        if(f[to[edge]]+val[edge]<s2)
        {
            s2=f[to[edge]]+val[edge];
            p2=to[edge];
        }
    }
    for(int edge=head[u];edge;edge=next[edge])
    if(to[edge]!=fa)
    {
        if(p1!=to[edge])
        fs[to[edge]]=min(fs[to[edge]],s1+val[edge]);else
        fs[to[edge]]=min(fs[to[edge]],s2+val[edge]);
        fs[to[edge]]=min(fs[to[edge]],gs[u]+val[edge]);
    }
    ll g1=0,g2=0;
    p1=0,p2=0;
    for(int edge=head[u];edge;edge=next[edge])
    if(to[edge]!=fa)
    {
        if(g[to[edge]]+val[edge]>g1)
        {
            g2=g1;
            g1=g[to[edge]]+val[edge];
            p2=p1;
            p1=to[edge];
        }else
        if(g[to[edge]]+val[edge]>g2)
        {
            g2=g[to[edge]]+val[edge];
            p2=to[edge];
        }
    }
    for(int edge=head[u];edge;edge=next[edge])
    if(to[edge]!=fa)
    {
        if(p1!=to[edge])
        gs[to[edge]]=max(gs[to[edge]],g1+val[edge]);else
        gs[to[edge]]=max(gs[to[edge]],g2+val[edge]);
        gs[to[edge]]=max(gs[to[edge]],fs[u]+val[edge]);
    }
    for(int edge=head[u];edge;edge=next[edge])
    if(to[edge]!=fa)
    {
        ans[to[edge]]=max(fs[to[edge]],f[to[edge]]);
        DP(to[edge],u);
    }
 } 
int main()
{
     scanf("%d",&n);
     for(int i=1;i<=n-1;i++)
     {
         int u,v,c;
         scanf("%d %d %d",&u,&v,&c);
         add_edge(u,v,c);
         add_edge(v,u,c);
    }
     for(int i=1;i<=n;i++) 
    {
        f[i]=gs[i]=0;
        g[i]=fs[i]=1e16;
    }
    dp(1,0);
       
    ans[1]=f[1];gs[1]=1e16;fs[1]=0;
    DP(1,0);
    for(int i=1;i<=n;i++)
    printf("%lld\\n",ans[i]);
}