乍一看，平衡树？

其实左偏树更好做啦\(qwq\)

每个节点都来棵左偏树维护最小值，\(dfs\)往上时合并一下，要是攻不下了就把根节点删掉，直到能攻下，

对了，攻下后值会变化怎么办？\(lazy\)标记一下，和线段树同理

My complete code:

#include
    
     #include
     
      using namespace std;typedef long long LL;const LL maxn=400000;struct node{    LL to,next;}dis[maxn];struct code{    LL val,lazy1,lazy2,dis;    LL son[2];}tree[maxn];LL n,m,num; LL scc[maxn],dep[maxn],ans[maxn],belong[maxn],tmp[maxn],k[maxn],def[maxn],change[maxn];LL head[maxn];inline void add(LL u,LL v){    dis[++num]=(node){v,head[u]}; head[u]=num;}inline void update(LL x){    LL lazy1=tree[x].lazy1,       lazy2=tree[x].lazy2,       son0=tree[x].son[0],       son1=tree[x].son[1];    if(son0){        tree[son0].val=tree[son0].val*lazy1+lazy2;        tree[son0].lazy1*=lazy1;        tree[son0].lazy2=tree[son0].lazy2*lazy1+lazy2;    }    if(son1){        tree[son1].val=tree[son1].val*lazy1+lazy2;        tree[son1].lazy1*=lazy1;        tree[son1].lazy2=tree[son1].lazy2*lazy1+lazy2;    }    tree[x].lazy1=1;    tree[x].lazy2=0;}LL merge(LL x,LL y){    if(!x || !y)        return x+y;    update(x);    update(y);    if(tree[x].val>tree[y].val)        swap(x,y);    tree[x].son[1]=merge(tree[x].son[1],y);    if(tree[tree[x].son[1]].dis>tree[tree[x].son[0]].dis)        swap(tree[x].son[1],tree[x].son[0]);    tree[x].dis=tree[tree[x].son[1]].dis+1;    return x;}void dfs(LL u,LL fa){    dep[u]=dep[fa]+1;    for(LL i=head[u];i;i=dis[i].next){        LL v=dis[i].to;        dfs(v,u);        if(!k[v]){            tree[belong[v]].val+=change[v];            tree[belong[v]].lazy2+=change[v];        }else{            tree[belong[v]].val*=change[v];            tree[belong[v]].lazy1*=change[v];            tree[belong[v]].lazy2*=change[v];        }        belong[u]=merge(belong[u],belong[v]);    }    while(belong[u] && tree[belong[u]].val