BZOJ3991: [SDOI2015]寻宝游戏

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=3991

可以发现答案是所有相邻藏宝点的lca和（第一个和最后一个也算相邻）

然后开个set维护一下插入和删除（插入inf和-inf就可以快速定位辣）。

#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<set>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 105000
#define inf int(1e9)
#define mm 1000000007
typedef long long ll;
using namespace std;
struct data{int obj,pre;ll c;
}e[maxn*2];
int bin[22],n,m,tot,idx;
int b[maxn],head[maxn],fa[maxn][22],dep[maxn],dfn[maxn],num[maxn];
ll d[maxn];
set<int> q;
ll read(){
    ll x=0,f=1; char ch=getchar();
    while (!isdigit(ch)){if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}   
void insert(int x,int y,ll z){
    e[++tot].obj=y; e[tot].pre=head[x]; e[tot].c=z; head[x]=tot; 
}
void dfs(int u){
    rep(i,1,20) if (dep[u]>=bin[i]) {
        fa[u][i]=fa[fa[u][i-1]][i-1]; 
    }
    dfn[u]=++idx; num[dfn[u]]=u;
    for (int j=head[u];j;j=e[j].pre){
        int v=e[j].obj;
        if (v!=fa[u][0]){
            d[v]=d[u]+e[j].c;
            dep[v]=dep[u]+1;
            fa[v][0]=u; 
            dfs(v);
        } 
    }
}
int lca(int x,int y){
    if (dep[x]<dep[y]) swap(x,y);
    int t=dep[x]-dep[y];
    rep(i,0,20) if (t&bin[i]) x=fa[x][i];
    down(i,20,0) if (fa[x][i]!=fa[y][i]) x=fa[x][i],y=fa[y][i];
    if (x==y) return x;
    return fa[x][0];
}
ll dis(int x,int y){
    int t=lca(x,y);
    return d[x]+d[y]-2*d[t];
}
int main(){
//  freopen("in.txt","r",stdin);
    bin[0]=1; rep(i,1,20) bin[i]=bin[i-1]*2;
    n=read(); m=read();
    rep(i,1,n-1){
        int x=read(),y=read();ll z=read();
        insert(x,y,z); insert(y,x,z);
    }
    dep[1]=1;
    dfs(1);
    q.insert(inf); q.insert(-inf);
    ll ans=0,tmp=0; int now,f;
    rep(i,1,m){
        int x=read(); now=dfn[x]; f=1;
        if (b[x]){
            f=-1; q.erase(now);
        } else q.insert(now);
        b[x]^=1;
        int l=*--q.lower_bound(now),r=*q.upper_bound(now);
        if (l!=-inf) ans+=f*dis(x,num[l]);
        if (r!=inf) ans+=f*dis(x,num[r]);
        if (l!=-inf&&r!=inf) ans-=f*dis(num[l],num[r]);
        if (q.size()!=2) tmp=dis(num[*q.upper_bound(-inf)],num[*--q.lower_bound(inf)]);
        printf("%lld\n",ans+tmp);
    }
     
    return 0;
}