CodeForces 383C

本文介绍了一种使用树状数组解决特定树形DP问题的方法。通过维护两个树状数组来分别记录奇数层与偶数层节点的值变化,从而实现对树形结构的有效更新与查询。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题目大意:

     有一颗树,两个操作:

                1.x,val   将点x加上val,x的子节点,如果与x的距离为偶数加上val,否则减去val。

                2.x         求点x的数值。

解决方法:

    建立两个树状数组,分别代表每次修改对于奇数层与偶数层的影响,然后求解是奇数层在奇数树状数组中求,偶数层在偶数树状数组中求。

我的代码:

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#define maxn 500000
using namespace std;
int tree1[maxn],tree2[maxn];
vector <int> po[maxn];
int l[maxn],r[maxn];
int d[maxn];
int num[maxn];
int n,m,totl;
int dfs(int v,int fa,int dd){
  d[v]=dd;
  l[v]=totl++;
  for (int i=0;i<po[v].size();i++)
    if (po[v][i]!=fa){
      dfs(po[v][i],v,1-dd);
    }
  r[v]=totl-1;
  return 0;
}
int add1(int x,int v){
  //cout<<x<<" "<<v<<" *"<<endl;
  for (int i=x;i<=n;i+=i&(-i))
    tree1[i]+=v;
  return 0;
}
int add2(int x,int v){
  //cout<<x<<" "<<v<<" *"<<endl;
  for (int i=x;i<=n;i+=i&(-i))
    tree2[i]+=v;
  return 0;
}
int sum1(int x){
  int ans=0;
  for (int i=x;i>0;i-=i&(-i))
    ans+=tree1[i];
  return ans;
}
int sum2(int x){
  int ans=0;
  for (int i=x;i>0;i-=i&(-i))
    ans+=tree2[i];
  return ans;
}
int main (){
  //freopen("test.in","r",stdin);
  while (~scanf("%d%d",&n,&m)){
    memset(tree1,0,sizeof(tree1));
    memset(tree2,0,sizeof(tree2));
    for (int i=1;i<=n;i++)
        scanf("%d",&num[i]),po[i].clear();
    for (int i=1;i<=n-1;i++){
        int a,b;scanf("%d%d",&a,&b);
        po[a].push_back(b);
        po[b].push_back(a);
    }
    totl=1;dfs(1,0,0);
    for (int i=1;i<=m;i++){
        int po,v;scanf("%d%d",&po,&v);
        if (po==1){
            int val;scanf("%d",&val);
            // cout<<"val v "<<val<<" "<<v<<" "<<l[v]<<" "<<r[v]<<endl;
            if (d[v]){
                add1(l[v],val);
                add1(r[v]+1,-1*val);
                add2(l[v],-1*val);
                add2(r[v]+1,val);
            }
            else {
                add1(l[v],-1*val);
                add1(r[v]+1,val);
                add2(l[v],val);
                add2(r[v]+1,-1*val);
            }
        }
        else {
            if (d[v]) cout<<sum1(l[v])+num[v]<<endl;
            else cout<<sum2(l[v])+num[v]<<endl;
        }
    }
  }
  return 0;
}


### Codeforces Problem 1332C Explanation The provided references pertain specifically to problem 742B on Codeforces rather than problem 1332C. For an accurate understanding and solution approach for problem 1332C, it's essential to refer directly to its description and constraints. However, based on general knowledge regarding competitive programming problems found on platforms like Codeforces: Problem 1332C typically involves algorithmic challenges that require efficient data structures or algorithms such as dynamic programming, graph theory, greedy algorithms, etc., depending upon the specific nature of the task described within this particular question[^6]. To provide a detailed explanation or demonstration concerning **Codeforces problem 1332C**, one would need direct access to the exact statement associated with this challenge since different tasks demand tailored strategies addressing their unique requirements. For obtaining precise details about problem 1332C including any sample inputs/outputs along with explanations or solutions, visiting the official Codeforces website and navigating to contest number 1332 followed by examining section C is recommended. ```python # Example pseudo-code structure often seen in solving competitive coding questions. def solve_problem_1332C(input_data): # Placeholder function body; actual logic depends heavily on the specifics of problem 1332C. processed_result = process_input(input_data) final_answer = compute_solution(processed_result) return final_answer input_example = "Example Input" print(solve_problem_1332C(input_example)) ```
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值