<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[chen_minghui的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/chen_minghui</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; chen_minghui]]></copyright><item><title><![CDATA[ubuntu自带输入法ibus 无法按数字键取词]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/80690821</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/80690821</guid><author>chen_minghui</author><pubDate>Thu, 14 Jun 2018 11:50:44 +0800</pubDate><description><![CDATA[
  rm -rf ~/.cache/ibus/libpinyin 
  有个人习惯词先备份user_bigram.db 
  重启 ibus restart
]]></description><category></category></item><item><title><![CDATA[”今日头条杯“湖北省赛 I. Five Day Couple 可持久化01字典树]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/80171155</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/80171155</guid><author>chen_minghui</author><pubDate>Wed, 02 May 2018 20:08:03 +0800</pubDate><description><![CDATA[题目链接https://www.nowcoder.com/acm/contest/104/H题意不带修改区间询问，每次给定一x，l，r。问x与a[l~r]中哪个数异或最大，输出这个最大值。思路对每一个前缀a[1~i] 建立一颗树，根结点为rt[i]。树中包含这个区间的所有数，结点中额外加个域cnt，记录有多少个数走到过这个节点。对于查询l,r。我们同时查询树rt[l-1]与树rt[r]。两颗树中的结]]></description><category></category></item><item><title><![CDATA[牛客练习赛8 E-集合中的质数 容斥原理]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78824517</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78824517</guid><author>chen_minghui</author><pubDate>Sun, 17 Dec 2017 12:51:59 +0800</pubDate><description><![CDATA[题目链接https://www.nowcoder.com/acm/contest/39/E#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<set>
#i]]></description><category></category></item><item><title><![CDATA[codeforces 896A Nephren gives a riddle dfs搜索]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78711490</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78711490</guid><author>chen_minghui</author><pubDate>Mon, 04 Dec 2017 17:01:46 +0800</pubDate><description><![CDATA[题目链接http://codeforces.com/problemset/problem/896/Acsrf_token=2d70d2ab176389681e17947374617961 
有意思的搜索#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int INF = ( 2e9 ) + 2;
const]]></description><category></category></item><item><title><![CDATA[codeforces  894E - Ralph and Mushrooms  tarjan缩点+数学+记忆化搜索找最长路]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78640033</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78640033</guid><author>chen_minghui</author><pubDate>Sun, 26 Nov 2017 21:21:29 +0800</pubDate><description><![CDATA[http://codeforces.com/problemset/problem/894/E 
发份代码记录一下，思路很简单，tarjan缩点之后计算点权值记忆化搜索找最长路。 
不过这个计算权值真的麻烦，推出的公式把long long 给爆了，debug了半天  
(╥╯^╰╥)#include<cstdio>
#include<iostream>
#include<algorithm>
#inc]]></description><category></category></item><item><title><![CDATA[HYNU 第二次周赛题解]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78633938</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78633938</guid><author>chen_minghui</author><pubDate>Sat, 25 Nov 2017 20:21:29 +0800</pubDate><description><![CDATA[问题 A: 小明A+B题目描述 
小明今年3岁了, 现在他已经能够认识100以内的非负整数, 并且能够进行100以内的非负整数的加法计算. 对于大于等于100的整数, 小明仅保留该数的最后两位进行计算, 如果计算结果大于等于100, 那么小明也仅保留计算结果的最后两位. 例如, 对于小明来说: 1) 1234和34是相等的 2) 35+80=15 给定非负整数A和B, 你的任务是代表小明计算出A+B]]></description><category></category></item><item><title><![CDATA[HDU 3695 Computer Virus on Planet Pandora AC自动机]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78584018</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78584018</guid><author>chen_minghui</author><pubDate>Mon, 20 Nov 2017 17:02:25 +0800</pubDate><description><![CDATA[题目链接http://acm.hdu.edu.cn/showproblem.php?pid=3695题意外星人的程序全是由大小写的英文字母组成，外星人也会写病毒程序。病毒程序也是大小写的英文字母组成，如果病毒是某个程序的子串，或者病毒的反转串是程序的子串，那么就说程序被该病毒感染。给出一个病毒及一个程序，问程序被之中的几个病毒感染了。思路将程序解析成字符串，正反在查询即可。。解析函数写错导致无限WA]]></description><category></category></item><item><title><![CDATA[HDU 5536 Chip Factory 带删除的01字典树]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78419685</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78419685</guid><author>chen_minghui</author><pubDate>Wed, 01 Nov 2017 22:35:15 +0800</pubDate><description><![CDATA[题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5536思路暴力查询#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#incl]]></description><category></category></item><item><title><![CDATA[POJ 3764 The xor-longest Path 01字典树+dfs]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78419672</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78419672</guid><author>chen_minghui</author><pubDate>Wed, 01 Nov 2017 22:33:51 +0800</pubDate><description><![CDATA[题目链接http://poj.org/problem?id=3764题意给定一颗树，要求找出一条路径，其边的权值异或和最大。思路求出各节点到根的异或和，假设两结点u，v到根的异或和为a[u],a[v] 那么u到v的异或和为a[u]^a[v] ,u，v的lca到根因为重复被去掉了。插入字典树中查询就行了。#include<cstdio>
#include<iostream>
#include<algo]]></description><category></category></item><item><title><![CDATA[HDU 4825 Xor Sum 01字典树模板]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78419624</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78419624</guid><author>chen_minghui</author><pubDate>Wed, 01 Nov 2017 22:29:51 +0800</pubDate><description><![CDATA[题目链接http://acm.hdu.edu.cn/showproblem.php?pid=4825思路01字符串模板题，将所有数从高位开始插入到树中。#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<string>
#include<queue>
#]]></description><category></category></item><item><title><![CDATA[POJ 1155 TELE 树形dp + 背包]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78376498</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78376498</guid><author>chen_minghui</author><pubDate>Sat, 28 Oct 2017 15:50:01 +0800</pubDate><description><![CDATA[题目链接http://poj.org/problem?id=1155题意一个信号传输网络是树形的。m个潜在用户，n-m个信号发射站。其中1号是电视台信号发电站。每个发射站之间连通需要一定的花费。用户都处在叶子节点处。且每个用户都会给定量的前补偿电视台。在使电视台不亏本的情况下，最多可以使多少用户连同。思路树形dp+分组背包 
用dp[u][j] 表示在u的子树中连同j个用户所能获得的最大价值 
如果]]></description><category></category></item><item><title><![CDATA[HDU 1712 ACboy needs your help 分组背包模板]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78376378</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78376378</guid><author>chen_minghui</author><pubDate>Sat, 28 Oct 2017 15:39:56 +0800</pubDate><description><![CDATA[题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1712题意有n门课，每门课所能获得的价值与在这门课上所花时间有关。用val[i][j] 表示在i课花费j天时间能获得的价值。给m天求最大价值。思路n门课，每门课都有m种选法，且只能选一种。分组背包模板。。#include<cstdio>
#include<iostream>
#include<algorit]]></description><category></category></item><item><title><![CDATA[HDU 4586 Information Disturbing 二分+树形dp]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78376310</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78376310</guid><author>chen_minghui</author><pubDate>Sat, 28 Oct 2017 15:33:34 +0800</pubDate><description><![CDATA[题目链接http://acm.hdu.edu.cn/showproblem.php?pid=3586题意有一个树形网络通信，1号结点是长官，叶子节点是前线通讯士兵，其他节点为中传递信息的士兵。为了打击这个通信网络，使其中任意一个前线士兵的通讯不能到达长官。利用一台机器可以切断士兵之间的通讯，破坏每对可以直接通讯的士兵都需要花费一个代价。在切断的边中，任意一条边的代价不能超过机器能力上限，且代价总和不]]></description><category></category></item><item><title><![CDATA[codeforces C. System Administrator 构造含割点的图]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78359331</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78359331</guid><author>chen_minghui</author><pubDate>Thu, 26 Oct 2017 22:07:24 +0800</pubDate><description><![CDATA[题目链接http://codeforces.com/problemset/problem/245/A题意构造一个含指定割点的连通图思路将v左边连一个，右边连n-2个，右边最大连边数为n+1+ (n-2)*(n-3)/2 ，然后n个点至少连n-1条边。#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#i]]></description><category></category></item><item><title><![CDATA[HDU 4714 Tree2cycle]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78359225</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78359225</guid><author>chen_minghui</author><pubDate>Thu, 26 Oct 2017 21:54:15 +0800</pubDate><description><![CDATA[题目链接https://vjudge.net/problem/HDU-4714题意给一个图，定义环为有n个结点，n条边的图。在图中每删和每加一条边需要花费，求用最少的花费将图变成环。思路思路比较难想到：看了题解才知道的。 
对于非根结点，假设其后代分支数为f，需要先将结点与其父亲断开，然后断开f-2个分支，将断开的分支再连成一条链，共需要花费1+2*(f-2)，在加上与上一条链的链接，总共需要2+2]]></description><category></category></item><item><title><![CDATA[POJ 3162 Walking Race 树的直径+线段树]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78358667</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78358667</guid><author>chen_minghui</author><pubDate>Thu, 26 Oct 2017 21:03:03 +0800</pubDate><description><![CDATA[题目链接http://poj.org/problem?id=3162题意给定一颗含n个结点的树，求每个点能走到的最远距离，挑选连续的k天 最长与最短之差不超过m，求最大k思路先求树的直径，然后线段树维护#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<]]></description><category></category></item><item><title><![CDATA[codeforces dfs序+线段树+bitset]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78357952</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78357952</guid><author>chen_minghui</author><pubDate>Thu, 26 Oct 2017 19:56:01 +0800</pubDate><description><![CDATA[题目链接http://codeforces.com/problemset/problem/620/E题意给定一个树，现在有60种颜色，要做如下操作 
1 v c：将v及其子树然成c颜色 
2 v：查询v及其子树中含有多少不同颜色思路用dfs序存储子树，bitset代表子树中含有的颜色。#include<cstdio>
#include<iostream>
#include<algorithm>
#i]]></description><category></category></item><item><title><![CDATA[POJ 3321 Apple Tree dfs序+线段树]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78357779</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78357779</guid><author>chen_minghui</author><pubDate>Thu, 26 Oct 2017 19:37:05 +0800</pubDate><description><![CDATA[题目链接http://poj.org/problem?id=3321题意一颗苹果树，以1为根，含有n个结点，每个节点上都会张出苹果。现在给出m个询问 
Q v代表修改v结点的状态，即有苹果就摘掉，没有苹果就让其长出苹果。 
C v代表查询以v为根的子树中，包含的苹果个数。思路用dfs序的连续区间可以代表一个子树，线段树叶子节点存储对应dfs序的结点苹果的有无。然后用线段树更新查询. 
用vector]]></description><category></category></item><item><title><![CDATA[POJ 3107 Godfather 求重心]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78357668</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78357668</guid><author>chen_minghui</author><pubDate>Thu, 26 Oct 2017 19:26:21 +0800</pubDate><description><![CDATA[题目链接http://poj.org/problem?id=3107#题意黑帮们之间的联系网是树状的，且将网络中的教父去除后，与去掉其他点相比，其最大子树是最小的。求出所有这些结点。思路求重心的模板题。。#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include]]></description><category></category></item><item><title><![CDATA[hdu 6201 transaction transaction transaction 树形dp]]></title><link>https://blog.youkuaiyun.com/chen_minghui/article/details/78225113</link><guid>https://blog.youkuaiyun.com/chen_minghui/article/details/78225113</guid><author>chen_minghui</author><pubDate>Fri, 13 Oct 2017 13:49:53 +0800</pubDate><description><![CDATA[题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=6201题意又是利用不同地方物价不同，赚得差价，这道题起点和终点可以任选，但是需要路费。从一颗树中，任选起点和终点，利润为差价减去路费。求最大利润思路用dp[u][0] :表示在u点买进时能够获得的最大价值 
用dp[u][1] :表示在u点卖出时能够获得的最大价值 
转移过程可以看代码#incl]]></description><category></category></item></channel></rss>