HDU 5296 Annoying problem LCA+树状数组
Annoying problem
Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 480 Accepted Submission(s): 146
Now there are two kinds of operation:
1 x: If the node x is not in the set S, add node x to the set S
2 x: If the node x is in the set S,delete node x from the set S
Now there is a annoying problem: In order to select a set of edges from tree after each operation which makes any two nodes in set S connected. What is the minimum of the sum of the selected edges’ weight ?
For each test:
The first line has 2 integer number n,q(0<n,q<=100000) describe the number of nodes and the number of operations.
The following n-1 lines each line has 3 integer number u,v,w describe that between node u and node v has an edge weight w.(1<=u,v<=n,1<=w<=100)
The following q lines each line has 2 integer number x,y describe one operation.(x=1 or 2,1<=y<=n)
The next q line represents the answer to each operation.
1
6 5
1 2 2
1 5 2
5 6 2
2 4 2
2 3 2
1 5
1 3
1 4
1 2
2 5
Case #1:
0
6
8
8
4
思路和代码:点击打开链接
HDU 5296 Annoying problem LCA+树状数组的更多相关文章
- HDU 3966 dfs序+LCA+树状数组
题目意思很明白: 给你一棵有n个节点的树,对树有下列操作: I c1 c2 k 意思是把从c1节点到c2节点路径上的点权值加上k D c1 c2 k 意思是把从c1节点到c2节点路径上的点权值减去k ...
- HDU 5296 Annoying problem (LCA,变形)
题意: 给一棵n个节点的树,再给q个操作,初始集合S为空,每个操作要在一个集合S中删除或增加某些点,输出每次操作后:要使得集合中任意两点互可达所耗最小需要多少权值.(记住只能利用原来给的树边.给的树边 ...
- hdu 6203 ping ping ping(LCA+树状数组)
hdu 6203 ping ping ping(LCA+树状数组) 题意:给一棵树,有m条路径,问至少删除多少个点使得这些路径都不连通 \(1 <= n <= 1e4\) \(1 < ...
- hdu 5517 Triple(二维树状数组)
Triple Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Sub ...
- HDU 5862 Counting Intersections(离散化+树状数组)
HDU 5862 Counting Intersections(离散化+树状数组) 题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=5862 D ...
- HDU 6203 ping ping ping(dfs序+LCA+树状数组)
http://acm.hdu.edu.cn/showproblem.php?pid=6203 题意: n+1 个点 n 条边的树(点标号 0 ~ n),有若干个点无法通行,导致 p 组 U V 无法连 ...
- BZOJ 4999: This Problem Is Too Simple! DFS序+LCA+树状数组+离线
Code: #include<bits/stdc++.h> #define setIO(s) freopen(s".in","r",stdin) , ...
- HDU 1394 Minimum Inversion Number ( 树状数组求逆序数 )
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1394 Minimum Inversion Number ...
- HDU 2689 Sort it【树状数组】
Sort it Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Sub ...
随机推荐
- storage存储对象和数组类型时候的问题
storage类型存储的类型为字符串,直接使用localstorage.setItem方法存储进去,取出来的时候数据是不能够使用的 解决方法: 先使用JSON.stringify方法转换成为字符串,然 ...
- bzoj 2792 [Poi2012]Well 二分+dp+two_pointer
题目大意 给出n个正整数X1,X2,...Xn,可以进行不超过m次操作,每次操作选择一个非零的Xi,并将它减一. 最终要求存在某个k满足Xk=0,并且z=max{|Xi - Xi+1|}最小. 输出最 ...
- 转 如何在C++中调用C程序
如何在C++中调用C程序? C++和C是两种完全不同的编译链接处理方式,如果直接在C++里面调用C函数,会找不到函数体,报链接错误.要解决这个问题,就要在 C++文件里面显示声明一下哪些函数是C写 ...
- AC日记——Propagating tree Codeforces 383c
C. Propagating tree time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- AC日记——[USACO07DEC]手链Charm Bracelet 洛谷 P2871
题目描述 Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like t ...
- js 路径改变时获取url参数
当我们在使用react或vue的router作路由跳转时,为了保持菜单与地址栏状态一致,我们可以使用window.onhashchange捕获#后面的变化 window.onhashchange = ...
- jquery 分页功能
<div class="wrapper"> <div class="row"> <div class="col-sm-1 ...
- Xamarin.Forms的基本页面和基本视图
Xamarin.Forms的基本页面和基本视图 在Xamarin.Forms中,每个App的界面都是一个页面Page.页面的种类有很多种.其中,最常见的页面就是内容页面ContentPage.项目 ...
- 几点iOS开发技巧
转自I'm Allen的博客 原文:iOS Programming Architecture and Design Guidelines 原文来自破船的分享 原文作者是开发界中知晓度相当高 ...
- C#使用反射机制获取类信息[转]
http://www.cnblogs.com/zhoufoxcn/archive/2006/10/31/2515873.html 1.用反射动态创建类实例,并调用其公有成员函数. //新建一个类库项目 ...