商务旅行

传送门
SPFA超时妥妥的！
N-1条边，树上找最短路，八成是LCA
用经过的点两两建立要查询的LCA，边权设为1.
就可以转化为小机房的树那个题目233

#include <cstdio>
#include <iostream>
using namespace std;
const int maxm=501000;
struct node{
    int net;
    int to;
    int cost;
    int lca;
    int ans;
};
bool vis[maxm];
int fat[maxm];
node edge[2][2*maxm];
int cnt[2],head[2][maxm];
int dis[maxm]; 
int t;
void add(int id,int x,int y,int c)
{
    edge[id][++cnt[id]].to=y;
    edge[id][cnt[id]].cost=c;
    edge[id][cnt[id]].net=head[id][x];
    head[id][x]=cnt[id];
}
int find(int x)
{
    if(fat[x]==x) return x;
    else return fat[x]=find(fat[x]);
}
void dfs(int x,int f,int sum)
{
    dis[x]+=dis[f]+sum;
    vis[x]=1;
    fat[x]=x;

    for(int K=head[0][x];K;K=edge[0][K].net)
    {
        int p=edge[0][K].to;
        if(vis[p]) continue;
        dfs(p,x,edge[0][K].cost);
        fat[p]=x;
    }

    for(int i=head[1][x];i;i=edge[1][i].net)
    {
        int p=edge[1][i].to;
        if(!vis[p]||edge[1][i].lca) continue;
        edge[1][i].lca=find(p); 
        int ans=dis[x]+dis[p]-2*dis[edge[1][i].lca];
        //printf("%d %d %d %d\n",x,p,edge[1][i].lca,ans);

        edge[1][i].ans=ans;
        t+=ans;
        if(i%2)
         edge[1][i+1].lca=edge[1][i].lca,edge[1][i+1].ans=ans;
        else
         edge[1][i-1].lca=edge[1][i].lca,edge[1][i-1].ans=ans;
    }
}
int main()
{
    int n,m;
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        add(0,x,y,1);
        add(0,y,x,1);
    }
    scanf("%d",&m);
    int last=1;
    //scanf("%d",&last);
    for(int i=1;i<=m;i++)
    {
        int x;
        scanf("%d",&x);
        add(1,last,x,0),add(1,x,last,0);
        last=x;
    }

    dfs(1,0,0);

    printf("%d",t);
}