P1552 [APIO2012]派遣

Posted 2020-11-13 SFWR

 

 

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因为限制了编号，所以直接逆序就是自底而上，然后就是树形DP合并节点了，然后用可并堆贪心的删除节点，每次更新节点答案

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#include<bits/stdc++.h>
using namespace std;
const int bow=100100;
int n,m,d[bow],lc[bow],rc[bow],f[bow],size[bow],sum[bow];
long long ans;
struct node{int sj,xs,ld;}num[bow];
int merge(int x,int y)
{
    if(!x||!y)return x+y;
    if(num[x].xs<num[y].xs)swap(x,y);
    rc[x]=merge(rc[x],y);
    if(d[lc[x]]<d[rc[x]])swap(lc[x],rc[x]);
    d[x]=d[rc[x]]+1;
    return x;
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        cin>>num[i].sj>>num[i].xs>>num[i].ld;
        ans=max(ans,1ll*num[i].ld);
        size[i]=1;sum[i]=num[i].xs;f[i]=i;
    }
    for(int i=n;i>=1;i--)
    {
        int ssj=num[i].sj;
        f[ssj]=merge(f[i],f[ssj]);
        size[ssj]+=size[i];sum[ssj]+=sum[i];
        while(sum[ssj]>m)
        {
            sum[ssj]-=num[f[ssj]].xs;
            f[ssj]=merge(lc[f[ssj]],rc[f[ssj]]);
            size[ssj]--;
        }
        ans=max(ans,1ll*num[ssj].ld*size[ssj]);
    }
    cout<<ans;
}