Codeforces 932.D Tree
2 seconds
512 megabytes
standard input
standard output
You are given a node of the tree with index 1 and with weight 0. Let cnt be the number of nodes in the tree at any instant (initially, cnt is set to 1). Support Q queries of following two types:
Add a new node (index cnt + 1) with weight W and add edge between node R and this node.
Output the maximum length of sequence of nodes which
- starts with R.
- Every node in the sequence is an ancestor of its predecessor.
- Sum of weight of nodes in sequence does not exceed X.
- For some nodes i, j that are consecutive in the sequence if i is an ancestor of j then w[i] ≥ w[j] and there should not exist a node k on simple path from i to j such that w[k] ≥ w[j]
The tree is rooted at node 1 at any instant.
Note that the queries are given in a modified way.
First line containing the number of queries Q (1 ≤ Q ≤ 400000).
Let last be the answer for previous query of type 2 (initially last equals 0).
Each of the next Q lines contains a query of following form:
- 1 p q (1 ≤ p, q ≤ 1018): This is query of first type where
and
. It is guaranteed that 1 ≤ R ≤ cnt and0 ≤ W ≤ 109.
- 2 p q (1 ≤ p, q ≤ 1018): This is query of second type where
and
. It is guaranteed that 1 ≤ R ≤ cntand 0 ≤ X ≤ 1015.
denotes bitwise XOR of a and b.
It is guaranteed that at least one query of type 2 exists.
Output the answer to each query of second type in separate line.
- 6
1 1 1
2 2 0
2 2 1
1 3 0
2 2 0
2 2 2
- 0
1
1
2
- 6
1 1 0
2 2 0
2 0 3
1 0 2
2 1 3
2 1 6
- 2
2
3
2
- 7
1 1 2
1 2 3
2 3 3
1 0 0
1 5 1
2 5 0
2 4 0
- 1
1
2
- 7
1 1 3
1 2 3
2 3 4
1 2 0
1 5 3
2 5 5
2 7 22
- 1
2
3
In the first example,
last = 0
- Query 1: 1 1 1, Node 2 with weight 1 is added to node 1.
- Query 2: 2 2 0, No sequence of nodes starting at 2 has weight less than or equal to 0. last = 0
- Query 3: 2 2 1, Answer is 1 as sequence will be {2}. last = 1
- Query 4: 1 2 1, Node 3 with weight 1 is added to node 2.
- Query 5: 2 3 1, Answer is 1 as sequence will be {3}. Node 2 cannot be added as sum of weights cannot be greater than 1. last = 1
- Query 6: 2 3 3, Answer is 2 as sequence will be {3, 2}. last = 2
题目大意:一棵树,每个点有点权,两种操作:1.新加一个点连接r,权值为w. 2.从一个点r开始往祖先上跳,每次跳到第一个值≥自身的祖先,将跳到的点的和加起来,不能大于w,问能跳几次.
分析:挺有意思的一道题.
暴力算法就是一个一个往上跳着找喽,在树上往上跳有一种常用的优化方法--倍增.在这道题里面可以倍增地跳到≥自身权值的点.
考虑怎么实现,fa数组就不能记录第2^i个祖先了,而要记录比自身权值大的第2^i个祖先.在加点的时候处理.可以发现,一旦处理出fa[i][0],就能够根据祖先节点的信息推出fa[i][j].
如何处理fa[i][0]?如果r的权值比新加的点i的权值大或相等,则fa[i][0] = r,否则从r开始往上跳,如果w[fa[r][j]] < w[i],则往上跳,最后fa[i][0] = fa[r][0].
因为最后要求和嘛,可以顺便维护一个sum数组,表示从i这个点跳到比i权值大的第2^j个祖先跳到的点的权值和为多少. 求出了这两个数组以后查询就很好办了,sum[i][j]是否≤w,是的话就往上跳,并且w -= sum[i][j].倍增的基础应用嘛.
想清楚如何加速往祖先跳的过程,以及倍增应该维护什么东西这道题就能解决了.
一些边界的值需要特殊考虑!
- #include <cstdio>
- #include <cstring>
- #include <iostream>
- #include <algorithm>
- using namespace std;
- typedef long long ll;
- const ll maxn = ,inf = 1e18;
- ll q,lastans,cnt = ,w[maxn];
- ll fa[maxn][],sum[maxn][];
- void add(ll x,ll v)
- {
- w[++cnt] = v;
- if (w[cnt] <= w[x])
- fa[cnt][] = x;
- else
- {
- int y = x;
- for (int i = ; i >= ; i--)
- {
- if (w[fa[y][i]] < w[cnt])
- y = fa[y][i];
- }
- fa[cnt][] = fa[y][];
- }
- if (fa[cnt][] == )
- sum[cnt][] = inf;
- else
- sum[cnt][] = w[fa[cnt][]];
- for (int i = ; i <= ; i++)
- {
- fa[cnt][i] = fa[fa[cnt][i - ]][i - ];
- if (fa[cnt][i] == )
- sum[cnt][i] = inf;
- else
- sum[cnt][i] = sum[cnt][i - ] + sum[fa[cnt][i - ]][i - ];
- }
- }
- ll query(ll x,ll v)
- {
- if (w[x] > v)
- return ;
- v -= w[x];
- ll res = ;
- for (int i = ; i >= ; i--)
- {
- if (v >= sum[x][i])
- {
- v -= sum[x][i];
- res += ( << i);
- x = fa[x][i];
- }
- }
- return res;
- }
- int main()
- {
- w[] = inf;
- for (int i = ; i <= ; i++)
- sum[][i] = inf;
- scanf("%I64d",&q);
- while (q--)
- {
- int id;
- ll a,b;
- scanf("%d",&id);
- scanf("%I64d%I64d",&a,&b);
- a ^= lastans;
- b ^= lastans;
- if (id == )
- add(a,b);
- else
- printf("%I64d\n",lastans = query(a,b));
- }
- }
Codeforces 932.D Tree的更多相关文章
- Problem - D - Codeforces Fix a Tree
Problem - D - Codeforces Fix a Tree 看完第一名的代码,顿然醒悟... 我可以把所有单独的点全部当成线,那么只有线和环. 如果全是线的话,直接线的条数-1,便是操作 ...
- Codeforces 765 E. Tree Folding
题目链接:http://codeforces.com/problemset/problem/765/E $DFS子$树进行$DP$ 大概分以下几种情况: 1.为叶子,直接返回. 2.长度不同的路径长度 ...
- Codeforces 932 E. Team Work(组合数学)
http://codeforces.com/contest/932/problem/E 题意: 可以看做 有n种小球,每种小球有无限个,先从中选出x种,再在这x种小球中任选k个小球的方案数 选出的 ...
- codeforces 570 D. Tree Requests 树状数组+dfs搜索序
链接:http://codeforces.com/problemset/problem/570/D D. Tree Requests time limit per test 2 seconds mem ...
- CodeForces 383C Propagating tree
Propagating tree Time Limit: 2000ms Memory Limit: 262144KB This problem will be judged on CodeForces ...
- 【19.77%】【codeforces 570D】Tree Requests
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- CodeForces - 274B Zero Tree
http://codeforces.com/problemset/problem/274/B 题目大意: 给定你一颗树,每个点上有权值. 现在你每次取出这颗树的一颗子树(即点集和边集均是原图的子集的连 ...
- Codeforces 343D Water Tree(DFS序 + 线段树)
题目大概说给一棵树,进行以下3个操作:把某结点为根的子树中各个结点值设为1.把某结点以及其各个祖先值设为0.询问某结点的值. 对于第一个操作就是经典的DFS序+线段树了.而对于第二个操作,考虑再维护一 ...
- codeforces 375D:Tree and Queries
Description You have a rooted tree consisting of n vertices. Each vertex of the tree has some color. ...
随机推荐
- 生成dataset的几种方式
1.常用的方式通过sparksession读取外部文件或者数据生成dataset(这里就不讲了) 注: 生成Row对象的方法提一下:RowFactory.create(x,y,z),取Row中的数据 ...
- access数据库频繁读取操作会出现 System.Data.OleDb.OleDbException 的异常解决
asp.net access数据库 本来想着打开一个access数据库连接后,不关闭,下次操作数据了,直接拿来用,谁知道连着测试64次后(大概这么多次),就会出现System.Data.OleDb.O ...
- Tempter of the Bone HDU 1010(DFS+剪枝)
Problem Description The doggie found a bone in an ancient maze, which fascinated him a lot. However, ...
- 20172326『Java程序设计』课程结对编程练习_四则运算第二周阶段总结
20172326『Java程序设计』课程结对编程练习_四则运算第二周阶段总结 小组成员 20172313 余坤澎 20172332 于欣月 20172326 康皓越 小组编程照片 设计思路 通过一个E ...
- 前端获取URL和SESSON中的值
.CS中代码 public ActionResult Index(string viewname, bool partial = false) { //获取URL中的 foreach (var key ...
- centos7环境下mysql安装
1.去官网下载合适的yum源安装包 https://dev.mysql.com/downloads/repo/yum/ 2.yum 本地安装 命令:yum localinstall mysql57-c ...
- RovingUI组件库-包含堆栈式通知提醒框(Toast)的小程序组件库
RovingUI是个人在开发小程序过程中将用到的组件集合而成的一个UI库,包含一些基本通用组件(按钮.栅格.通用样式.徽标.通知和面包屑). 源起得归于我在开发中没有找到现成的堆栈式提醒框(比如ant ...
- 读着读着《构建之法》(Build To Win) 越精神的白雪儿的思考
哲学家的宗旨是:我思,故我在 科学家的宗旨是:我发现,故我在 工程师的宗旨是:我构建,故我在 ——<工程学--无尽的前沿> 序言:珍惜角色“人”,注重实践“物” <构建之法>, ...
- 3dContactPointAnnotationTool开发日志(三十)
在vs2017里生成opencv时遇到了无法打开python27_d.lib的问题,具体解决请看这个,不过我用的是方法2,python37_d.lib找不到同理. Windows下可以用的op ...
- rfid工作原理
RFID的工作原理是:标签进入磁场后,如果接收到阅读器发出的特殊射频信号,就能凭借感应电流所获得的能量发送出存储在芯片中的产品信息(即Passive Tag,无源标签或被动标签),或者主动发送某一频率 ...