「JLOI2015」城池攻占 可并堆

传送门

分析

如果直接暴力枚举的话肯定会超时
我们可以从下往上遍历，维护一个小根堆
每次到达一个节点把战败的骑士扔出去
剩下的再继续向上合并，注意要维护一下其实的战斗力
可以像线段树那样用一个lazy标记

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn=300005;
typedef long long ll;
ll head[maxn],tot=1;
struct asd{
    ll from,to,next,ty;
    ll val;
}b[maxn];
void ad(ll aa,ll bb,ll cc,ll dd){
    b[tot].from=aa;
    b[tot].to=bb;
    b[tot].ty=cc;
    b[tot].val=dd;
    b[tot].next=head[aa];
    head[aa]=tot++;
}
ll bjc[maxn],bjj[maxn],gjl[maxn];
ll lch[maxn],rch[maxn];
void push_down(ll xx){
    if(xx==0) return;
    if(bjc[xx]!=1){
        gjl[lch[xx]]*=bjc[xx],bjj[lch[xx]]*=bjc[xx],bjc[lch[xx]]*=bjc[xx];
        gjl[rch[xx]]*=bjc[xx],bjj[rch[xx]]*=bjc[xx],bjc[rch[xx]]*=bjc[xx];
        bjc[xx]=1;
    }
    if(bjj[xx]!=0){
        gjl[lch[xx]]+=bjj[xx],bjj[lch[xx]]+=bjj[xx];
        gjl[rch[xx]]+=bjj[xx],bjj[rch[xx]]+=bjj[xx];
        bjj[xx]=0;
    }
}
ll d[maxn],dep[maxn],rt[maxn];
ll h[maxn];
ll bing(ll xx,ll yy){
    if(!xx) return yy;
    if(!yy) return xx;
    push_down(xx),push_down(yy);
    if(gjl[xx]>gjl[yy]) swap(xx,yy);
    rch[xx]=bing(rch[xx],yy);
    if(d[lch[xx]]<d[rch[xx]]) swap(lch[xx],rch[xx]);
    d[xx]=d[rch[xx]]+1;
    return xx;
}
ll killl[maxn],atk[maxn];
ll cnt=0;
void dfs(ll xx){
    for(ll i=head[xx];i!=-1;i=b[i].next){
        ll u=b[i].to;
        dep[u]=dep[xx]+1,dfs(u);
        if(b[i].ty) gjl[rt[u]]*=b[i].val,bjj[rt[u]]*=b[i].val,bjc[rt[u]]*=b[i].val;
        else gjl[rt[u]]+=b[i].val,bjj[rt[u]]+=b[i].val;
        rt[xx]=bing(rt[xx],rt[u]);
    }
    while(rt[xx] && gjl[rt[xx]]<h[xx]){
        killl[xx]++,atk[rt[xx]]=xx,push_down(rt[xx]),rt[xx]=bing(lch[rt[xx]],rch[rt[xx]]);
    }
}
ll jl[maxn];
int main(){
    memset(head,-1,sizeof(head));
    ll n,m;
    scanf("%lld%lld",&n,&m);
    for(ll i=1;i<=n;i++) scanf("%lld",&h[i]);
    for(ll i=2;i<=n;i++){
        ll aa;
        ll bb,cc;
        scanf("%lld%lld%lld",&aa,&bb,&cc);
        ad(aa,i,bb,cc);
    }
    for(ll i=1;i<=m;i++){
        bjc[i]=1;
        ll bb;
        scanf("%lld%lld",&gjl[i],&bb);
        jl[i]=bb;
        rt[bb]=bing(rt[bb],i);
    }
    d[0]=-1;
    dep[1]=1;
    dfs(1);
    for(ll i=1;i<=n;i++) printf("%lld\n",killl[i]);
    for(ll i=1;i<=n;i++) printf("%lld\n",dep[jl[i]]-dep[atk[i]]);
    return 0;
}