思路 恶心人的题目 还是类似永无乡一题的Treap启发式合并思路 但是由于加边变成了删边 所以应该离线后倒序处理 数组要开够 代码 #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #include <set> #include <stack> using namespace std; int Nodecnt=0,root[250000…

Graph and Queries Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) [Problem Description] You are given an undirected graph with N vertexes and M edges. Every vertex in this graph has an integer value assigned to it…

[la P5031&hdu P3726] Graph and Queries Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description You are given an undirected graph with N vertexes and M edges. Every vertex in this graph has an integer v…

Graph and Queries Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1467    Accepted Submission(s): 301 Problem Description You are given an undirected graph with N vertexes and M edges. Every ve…

题目来源:HDU 3726 Graph and Queries 题意:见白书 思路:刚学treap 參考白皮书 #include <cstdio> #include <cstring> #include <cstdlib> using namespace std; struct Node { Node *ch[2]; int r; int v; int s; Node(int v): v(v) { ch[0] = ch[1] = NULL; r = rand(); s…

\[ \text{Preface} \] 算是一道思维难度稍易,代码难度稍难的题吧. \[ \text{Description} \] 给出一张 \(n\) 个点,\(m\) 条边的图,点带权.需要支持三个操作: D x 删掉编号为 \(x\) 的边 Q x k 查询与节点 \(x\) 联通的所有节点中,点权第 \(k\) 大节点的点权 C x v 将节点 \(x\) 点权改为 \(v\) 多组数据,每组数据最终需要输出所有查询的平均值 ( 保留 6 位 ) ,没有强制在线. \[ \text{…

Description You are given an undirected graph with N vertexes and M edges. Every vertex in this graph has an integer value assigned to it at the beginning. You're also given a sequence of operations and you need to process them as requested. Here's a…

反向操作,先求出最终状态,再反向操作. 然后就是Treap 的合并,求第K大值. #include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<map> #include<queue> #include<vector> #inc…

这题写起来真累.. 名次树就是多了一个附加信息记录以该节点为根的树的总结点的个数,由于BST的性质再根据这个附加信息,我们可以很容易找到这棵树中第k大的值是多少. 所以在这道题中用一棵名次树来维护一个连通分量. 由于图中添边比较方便,用并查集来表示连通分量就好了,但是删边不太容易实现. 所以,先把所有的边删去,然后逆序执行命令.当然,C命令也要发生一些变化,比如说顺序的情况是从a变成b,那么逆序执行的话应该就是从b变成a. 最后两棵树的合并就是启发式合并,把节点数少的数并到节点数多的数里去. #…

题意:给你个点m条边的无向图,每个节点都有一个整数权值.你的任务是执行一系列操作.操作分为3种... 思路:本题一点要逆向来做,正向每次如果删边,复杂度太高.逆向到一定顺序的时候添加一条边更容易.详见算法指南P235. #include<cstdlib> struct Node { Node *ch[]; // 左右子树 int r; // 随机优先级 int v; // 值 int s; // 结点总数 Node(int v):v(v) { ch[] = ch[] = NULL; r = r…