Problem 给你一张图,点的权值,边和几个操作: D x: 删除第x条边 Q x y: 询问包含x的联通块中权值第y大的权值 C x y: 将x这个点的权值改为y Solution 一看就要离线处理,把所有操作都倒过来 然后删除操作变为加边操作 Notice 记得: 是改完以后再把点一个一个加入Treap中!! Code 非旋转Treap #pragma GCC optimize(2) #include<cmath> #include<cstdio> #include<c…

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…

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…

[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…

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…

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…

Treap的基础题目,Treap是个挺不错的数据结构. /* */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <deque> #include <algorithm> #include…