CodeForces - 1099F：Cookies （线段树）

Mitya and Vasya are playing an interesting game. They have a rooted tree with $n$

There are some cookies in every vertex of the tree: there are i

$n$

with its parent.

Mitya and Vasya take turns playing, Mitya goes first.

Mitya moves the chip from the vertex, where the chip is located, to one of its children.
Vasya can remove an edge from the vertex, where the chip is located, to one of its children. Vasya can also decide to skip his turn.

$CodeForces - 1099F：Cookies （线段树）$

Mitya can stop the game at any his turn. Once he stops the game, he moves the chip up to the root, eating some cookies along his way. Mitya can decide how many cookies he would like to eat in every vertex on his way. The total time spent on descend, ascend and eating cookies should not exceed T

$n$

time).

Find out what is the maximum number of cookies Mitya can eat, regardless of Vasya's actions.

$n$

The first line contains two integers n

$n$

The second line contains n

$n$

Each of the following 1

$n$

Output a single integer — maximum number of cookies Mitya can eat.

$n$

Input

Output

Input

Output

$n$

In the first example test case, Mitya can start by moving the chip to vertex 2

$n$

cookies. Below you can find the detailed description of the moves:

Mitya moves chip to vertex 2

.
Vasya removes edge to vertex 4
.
Mitya moves chip to vertex 5
.
Since vertex 5
has no children, Vasya does not remove any edges.
Mitya stops the game and moves the chip towards the root, eating cookies along the way (7

$n$

Mitya spend 0

$n$

题意：给定带权树，每个节点有一些a[i]个吃的，每个需要时间t[i]，现在让你找一条路线，可以吃到最多的吃的。

思路：对于当前路线1-u，减去路上花费的时间，剩下的时间T-time用来吃东西，我们肯定是按照时间t排序后从小到大吃，这里不嫩想到用二分+前缀和来做。

线段树可以很好的解决这个问题。（参考了WXK的代码，非常的漂亮，orz

题目还有博弈的限制，其实就是除了1号节点找第二大即可。

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=4000010;
ll sum[maxn],num[maxn],Mx[maxn][2],ans[maxn];int t[maxn],x[maxn];
int Laxt[maxn],Next[maxn],To[maxn],Len[maxn<<1],cnt;
void add(int u,int v,int l){
    Next[++cnt]=Laxt[u]; Laxt[u]=cnt; To[cnt]=v; Len[cnt]=l;
}
void update(int Now,int L,int R,int pos,int val)
{
    sum[Now]+=(ll)val*pos; num[Now]+=val;
    if(L==R) return ;int Mid=(L+R)>>1;
    if(pos<=Mid) update(Now<<1,L,Mid,pos,val);
    else update(Now<<1|1,Mid+1,R,pos,val);
}
ll query(int Now,int L,int R,ll T)
{
    if(sum[Now]<=T) return num[Now];
    if(L==R) return T/L;
    int Mid=(L+R)>>1;
    if(sum[Now<<1]<=T) return num[Now<<1]+query(Now<<1|1,Mid+1,R,T-sum[Now<<1]);
    return query(Now<<1,L,Mid,T);
}
void dfs(int u,ll T)
{
    if(T<=0) return ;
    update(1,1,1000000,t[u],x[u]);
    ans[u]=query(1,1,1000000,T);
    for(int i=Laxt[u];i;i=Next[i]){
        dfs(To[i],T-Len[i]*2);
        ll z=ans[To[i]];
        if(z>Mx[u][1]) swap(z,Mx[u][1]);
        if(z>Mx[u][0]) swap(z,Mx[u][0]);
    }
    if(u!=1) ans[u]=max(Mx[u][0],ans[u]);
    else ans[u]=max(Mx[u][1],ans[u]);
    update(1,1,1000000,t[u],-x[u]);
}
int main()
{
    int N,p,l; ll T;
    scanf("%d%lld",&N,&T);
    rep(i,1,N) scanf("%d",&x[i]);
    rep(i,1,N) scanf("%d",&t[i]);
    rep(i,2,N){
        scanf("%d%d",&p,&l);
        add(p,i,l);
    }
    dfs(1,T);
    printf("%lld\n",ans[1]);
    return 0;
}