<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Strive]]></title><description><![CDATA[努力不断努力]]></description><link>https://blog.youkuaiyun.com/HAHAAC</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; HAHAAC]]></copyright><item><title><![CDATA[最小树形图]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/78748885</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/78748885</guid><author>HAHAAC</author><pubDate>Fri, 08 Dec 2017 10:28:10 +0800</pubDate><description><![CDATA[/*本题为不是固定根的最小树形图，我们可以虚拟出一根来，然后在把这个根跟每个点相连，相连的点可以设为无穷大，或者设为所有边和大一点，比如为r，然后就可以利用最小树形图进行计算了，计算出的结果减去r,如果比r还大就可以认为通过这个虚拟节点我们连过原图中两个点，即原图是不连通的，我们就可以认为不存在最小树形图。关于输出最小根也挺简单，在找最小入弧时，如果这条弧的起点是虚拟根，那么这条弧的终点就是要求的]]></description><category></category></item><item><title><![CDATA[HDU 2121 无固定点的最小树形图]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53218147</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53218147</guid><author>HAHAAC</author><pubDate>Fri, 18 Nov 2016 16:15:27 +0800</pubDate><description><![CDATA[/*
本题为不是固定根的最小树形图，我们可以虚拟出一根来，然后在把这个根跟每个点相连，相连的点可以设为无穷大，或者设为所有边和大一点，比如为r，然后就可以利用最小树形图进行计算了，计算出的结果减去r,如果比r还大就可以认为通过这个虚拟节点我们连过原图中两个点，即原图是不连通的，我们就可以认为不存在最小树形图。关于输出最小根也挺简单，在找最小入弧时，如果这条弧的起点是虚拟根，那么这条弧的终点就是要求]]></description><category></category></item><item><title><![CDATA[poj 3264 Command Network(最小树形图)]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53218073</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53218073</guid><author>HAHAAC</author><pubDate>Fri, 18 Nov 2016 16:13:18 +0800</pubDate><description><![CDATA[有一副有向图，要求从根节点起能访问到所有的结点，且路径最短。即求最小树形图。
算法位朱刘算法，
裸的最小树形图，用朱—刘算法解决，具体实现过程如下：算法一开始先判断从固定根开始是否可达所有原图中的点，若不可，则一定不存在最小树形图。这一步是一个很随便的搜索，写多搓都行，不加废话。第二步，遍历所有的边，从中找出除根结点外各点的最小入边，累加权值，构成新图。接着判断该图是否存在环。若不存在，则该图]]></description><category></category></item><item><title><![CDATA[poj 1679 The Unique MST （[kuangbin带你飞]专题八 生成树 ）]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53175985</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53175985</guid><author>HAHAAC</author><pubDate>Tue, 15 Nov 2016 20:00:53 +0800</pubDate><description><![CDATA[题目大意：给你一个无向图，问如果这个无向图的最小生成树是唯一的就输出最小生成树的权值，否则输出Not Unique!
解题思路，求次小生成树，如果次小生成树中有何最小的权值一样的，则生成树不唯一，否则唯一用Kruskal，求最小生成树，然后枚举其中一个边去掉以后（n-1条），再求最小生成树，然后和原来的权值比较，


#include
#include
#include
using nam]]></description><category></category></item><item><title><![CDATA[poj 3026 Borg Maze]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53168684</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53168684</guid><author>HAHAAC</author><pubDate>Tue, 15 Nov 2016 10:24:20 +0800</pubDate><description><![CDATA[题目大意：s会分裂，然后从s到A的最小的步是多少。
解题思路：先bfs算出A或者S到任何点的距离，然后最求最小生成树，用dij或者prim都可以




#include
#include
#include
#include
#include
using namespace std;
const int maxn = 100+10;
char mp[maxn][maxn];
int c]]></description><category></category></item><item><title><![CDATA[POJ 1797  Heavy Transportation]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53116725</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53116725</guid><author>HAHAAC</author><pubDate>Thu, 10 Nov 2016 16:29:01 +0800</pubDate><description><![CDATA[Description

Background 
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from t]]></description><category></category></item><item><title><![CDATA[poj 3268 Silver Cow Party ([kuangbin带你飞]专题四 最短路练习)]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53116426</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53116426</guid><author>HAHAAC</author><pubDate>Thu, 10 Nov 2016 15:49:07 +0800</pubDate><description><![CDATA[Description


One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirect]]></description><category></category></item><item><title><![CDATA[poj 1860 Currency Exchange ([kuangbin带你飞]专题四 最短路练习)]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53116300</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53116300</guid><author>HAHAAC</author><pubDate>Thu, 10 Nov 2016 15:40:12 +0800</pubDate><description><![CDATA[Description

Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these curren]]></description><category></category></item><item><title><![CDATA[wfu第一次周赛代码]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53069595</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53069595</guid><author>HAHAAC</author><pubDate>Mon, 07 Nov 2016 19:55:11 +0800</pubDate><description><![CDATA[第一题和第二题都没有代码，，，，，，不会自己想办法
第三题：
#include
#include

int main(){
    int a,b,i;
    const int maxn = 10;
    int B[maxn];
    while(scanf("%d%d",&a,&b)!=EOF){
        int sum=0;
        int j=0;
      ]]></description><category></category></item><item><title><![CDATA[poj 2240 Arbitrage （[kuangbin带你飞]专题四 最短路练习）]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53010481</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53010481</guid><author>HAHAAC</author><pubDate>Wed, 02 Nov 2016 18:33:04 +0800</pubDate><description><![CDATA[Description

Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar ]]></description><category></category></item><item><title><![CDATA[poj 3660 Cow Contest ([kuangbin带你飞]专题四 最短路练习)]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/53008598</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/53008598</guid><author>HAHAAC</author><pubDate>Wed, 02 Nov 2016 15:45:23 +0800</pubDate><description><![CDATA[Description


N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill ]]></description><category></category></item><item><title><![CDATA[poj 1502 MPI Maelstrom([kuangbin带你飞]专题四 最短路练习)]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/52997691</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/52997691</guid><author>HAHAAC</author><pubDate>Tue, 01 Nov 2016 17:10:40 +0800</pubDate><description><![CDATA[Description

BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee']]></description><category></category></item><item><title><![CDATA[poj 3259 Wormholes ([kuangbin带你飞]专题四 最短路练习)]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/52994098</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/52994098</guid><author>HAHAAC</author><pubDate>Tue, 01 Nov 2016 11:28:23 +0800</pubDate><description><![CDATA[Description


While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination]]></description><category></category></item><item><title><![CDATA[HDU 5135  Little Zu Chongzhi's Triangles （2014 ACM-ICPC广州站）]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/52904561</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/52904561</guid><author>HAHAAC</author><pubDate>Sun, 23 Oct 2016 21:26:36 +0800</pubDate><description><![CDATA[题目链接：题目
题目大意：给你N的数，求能组成所有三角行的面积和的最大值
解题思路：状态压缩dp，还没接触过dp，，，，感觉好难啊，，
我也不会写，说下思路把，就是枚举所有的状态，，其中三个一组，组成三角行，并且边不能重复用


#include
#include
#include
#include 
using namespace std;
const int maxn = 20;
]]></description><category></category></item><item><title><![CDATA[HDU 5918 Sequence I （2016-ccpc-长春）]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/52863530</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/52863530</guid><author>HAHAAC</author><pubDate>Wed, 19 Oct 2016 21:26:05 +0800</pubDate><description><![CDATA[题目连接：题目
题目大意：给你数列a，b，求在数列a中找出有几个间隔为p的数列b
解题思路：直接暴力，复杂度n*m，
#include
#include
using namespace std;
const int maxn = 1e6+5;
int a[maxn],b[maxn];
int n,m,p;
int sovle(int a[],int b[],int k){
	int sum=]]></description><category></category></item><item><title><![CDATA[HDU 5916 Harmonic Value Description （2016-ccpc-长春）]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/52863421</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/52863421</guid><author>HAHAAC</author><pubDate>Wed, 19 Oct 2016 21:22:07 +0800</pubDate><description><![CDATA[题目连接：题目
题目大意：
给出N，K，给你1~n个数，在所有全排列中相邻的两个数的gcd的和第K大的排列，排列有多种，输出其中一种
解题思路：
设，p是左右gcd的和，则p是从n-1，开始的连续数字
当k==1时就是输出从1~n
当k时偶数时，把k和2*k提到最前面，因为k和2*k的最大公约数时k，别的数都不影响，他们的最大公约数还是1，所以p时k+n-2，也就是从n-1开始数的第K]]></description><category></category></item><item><title><![CDATA[HDU 5914 Triangle （2016-ccpc-长春）]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/52862940</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/52862940</guid><author>HAHAAC</author><pubDate>Wed, 19 Oct 2016 20:31:58 +0800</pubDate><description><![CDATA[题目链接：题目
题目大意：给你1~n(1


解题思路：把1~n中不是斐波那契数列中数全部去掉就行了，剩下的数一定不能组成三角形


#include
#include
using namespace std;
int a[]={0,0,0,0,1,1,2,3,3,4,5,6,7,7,8,9,10,11,12,13,14};
int main(){
	int t;
	int ca=1]]></description><category></category></item><item><title><![CDATA[HDU 5912 Fraction (2016-ccpc-长春)]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/52862906</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/52862906</guid><author>HAHAAC</author><pubDate>Wed, 19 Oct 2016 20:26:19 +0800</pubDate><description><![CDATA[题目连接：题目
求公式结果，数据量比较小，递归一下就行了，和某年的蓝桥杯一个题一样
#include
#include
#include
#include
using namespace std;
const int maxn = 10;
int n;
int a[maxn],b[maxn];
int gcd(int x,int y){
    if(x==0) return y;
    e]]></description><category></category></item><item><title><![CDATA[HDU 5512 Pagodas（2015-ICPC-沈阳站）]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/52837804</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/52837804</guid><author>HAHAAC</author><pubDate>Mon, 17 Oct 2016 12:44:05 +0800</pubDate><description><![CDATA[题目连接：题目
题目大意：有n个庙经过长时间风吹雨打需要修补，只有两座（被标记为a，b）完好无损不需要修补，有两个和尚轮流去修补这n-2个庙，每个和尚每次只能修补一个庙标记为i，并要求i满足i=j+k或者i=j-k，每个庙只能被修建一次；其中j和k代表已经修建好的庙，Yuwgna先开始，问最后谁不能修建谁输；


解题思路
博弈论，看剩下的庙是不是GCD(a,b)的倍数，偶数倍是Yuwb]]></description><category></category></item><item><title><![CDATA[HDU 5573 Binary Tree（2015-ICPC-上海赛区）]]></title><link>https://blog.youkuaiyun.com/HAHAAC/article/details/52837710</link><guid>https://blog.youkuaiyun.com/HAHAAC/article/details/52837710</guid><author>HAHAAC</author><pubDate>Mon, 17 Oct 2016 12:18:24 +0800</pubDate><description><![CDATA[题目连接：题目
大意：
有一颗完全二叉树，第一个节点是1，他的左儿子就是i x 2，右儿子是i x 2+1
然后让你找到一个路径，使得通过加减恰好向下走k步之后，权值和为n，
解题思路：


构造题

首先我们可以分析得到，所有答案都可以通过走 1,2,4,8,16.....来得到

为什么？因为n

如果是奇数，那么我们最后一步就选左儿子，否则选择右儿子

路径问题]]></description><category></category></item></channel></rss>