http://acm.hdu.edu.cn/showproblem.php?pid=5521 题目大意:A,B两个人分别在1和n区.给出区之间有联系的图以及到达所需时间.求两个人见面最短时间以及在哪个区碰面(可有多个).多个区被当成一个集合(set),集合中的点互相到达时间相同,并且一定能互达. 思路:在于边的数量庞大.需要着手解决的问题是如何建图使得边的数目减小. 题目的特殊点在于:每个区域内部的点点距离是相同的. 所以,构造虚拟结点,解决边数过多的问题. (n*(n-1))/2 ------…

Meeting Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 6865    Accepted Submission(s): 2085 Problem Description Bessie and her friend Elsie decide to have a meeting. However, after Farmer Jo…

Meeting Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 3361    Accepted Submission(s): 1073 Problem Description Bessie and her friend Elsie decide to have a meeting. However, after Farmer Jo…

今天做的沈阳站重现赛,自己还是太水,只做出两道签到题,另外两道看懂题意了,但是也没能做出来. 1. Thickest Burger Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 0    Accepted Submission(s): 0 Problem Description ACM ICPC is launching a thic…

Bazinga Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 214    Accepted Submission(s): 89 Problem Description Ladies and gentlemen, please sit up straight.Don't tilt your head. I'm serious.For n…

Pagodas Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 70    Accepted Submission(s): 62 Problem Description n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yun…

Recursive sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 249    Accepted Submission(s): 140 Problem Description Farmer John likes to play mathematics games with his N cows. Recently, t…

Counting Cliques Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 539    Accepted Submission(s): 204 Problem Description A clique is a complete graph, in which there is an edge between every pair…

Thickest Burger Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 63    Accepted Submission(s): 60 Problem Description ACM ICPC is launching a thick burger. The thickness (or the height) of a piec…

Relative atomic mass Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 66    Accepted Submission(s): 59 Problem Description Relative atomic mass is a dimensionless physical quantity, the ratio of…

Pagodas Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1282    Accepted Submission(s): 902 Problem Description n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the…

Bazinga Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3509    Accepted Submission(s): 1122 Problem Description Ladies and gentlemen, please sit up straight.Don't tilt your head. I'm serious.Fo…

Rabbits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1193    Accepted Submission(s): 628 Problem Description Here N (N ≥ 3) rabbits are playing by the river. They are playing on a number li…

整理代码... Little Boxes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2304    Accepted Submission(s): 818 Problem Description Little boxes on the hillside.Little boxes made of ticky-tacky.Littl…

Minimum Cut[贪心]2015沈阳online 题意:割最少的边使得图不连通,并且割掉的边中有且仅有一条是生成树的边. 首先,我们选择一条树中的边进行切割,此时仅考虑树上的边集,有两种情况:1.树被分为两个结点数大于1的子树2.树被分为一个子树和一个单独结点. 如果选择第一种情况,那么还要割掉其中一颗子树中的结点所涉及到的所有非树中边,这种情况下,如果重新选择,选择子树上的一个叶子结点,那么只需要割掉叶子结点于该树所连的边和该结点涉及到的非树中边,所需要割的边数一定小于第一种情形. 代码…

题意:有\(n\)个点,\(m\)个集合,集合\(E_i\)中的点都与集合中的其它点有一条边权为\(t_i\)的边,现在问第\(1\)个点和第\(n\)个点到某个点的路径最短,输出最短路径和目标点,如果不满足条件则输出\(Evil John\). 题解:题目所给的边数关系太复杂了,我们可以让每个集合中的所有点都与一个虚拟节点连边,而这些点两两却不连,然后再去找\(1\)个和第\(n\)个点的最短路径,不难发现,最终得到的路径为\(dis[i]/2\),所以我们只要用虚拟节点建边然后跑两次dijk…

题目链接  2015 ACM-ICPC Shenyang Problem A Problem B Problem C Problem D 签到题,所有gcd的倍数都可以被写出来. 那么判断一下这类数的个数的奇偶性就可以了. Problem E Problem F Problem G Problem H Problem I 把三元组分类之后发现符合题意条件的只可能是那些第一维是最大值的那些三元组. 记录一下每个可能符合条件的三元组及其个数(可能同一个会有多个). 然后先对第一维计数排序,然后对二三…

M - Meeting Time Limit:6000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit cid=113817#status//M/0" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" style="display:inline-block;pad…

题目链接: Meeting Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2024    Accepted Submission(s): 628 Problem Description Bessie and her friend Elsie decide to have a meeting. However, after Farm…

版权所有所有:没有马缰绳chhuach(CSDN博客源).转载请注明出处. 禁止www.haogongju.net转载. 特此声明 一.开篇: 9月底,找工作接近尾声,笔者主要经历了2015年南京站百度.阿里.腾讯.美团.趋势科技.大众点评.华为的笔试.面试.当中拿到百度.美团.趋势科技.华为的软件研发offer. 找工作是幸苦的,笔者曾一天来回跑面百度.大众点评.趋势科技三家公司. 一天面试4面是常见的.一大早出门,有时候连午饭都顾不上吃,回到学校就5点了.吃过晚饭,一天就过了.中间有非常多打…

A - Pattern String 留坑. B - Bazinga 题意:找一个最大的i,使得前i - 1个字符串中至少不是它的子串 思路:暴力找,如果有一个串已经符合条件,就不用往上更新 #include <bits/stdc++.h> using namespace std; #define N 510 int t, n; int vis[N]; string s[N]; int main() { ios::sync_with_stdio(false); int t; cin >&…

首发于QQ空间和知乎,我在这里也更一下.   前言 以前高中搞竞赛的时候,经常看到神犇出去比赛或者训练之后写游记什么的,感觉萌萌哒.但是由于太弱,就没什么心情好写.现在虽然还是很弱,但是抱着享受的心情来沈阳玩一圈,总结一下还是可以的……  day -n  某天去数院找学长搞请假条,被无情地奶了一口:“你们要拿金回来啊!”此处应有一个巨大的flag……为滚粗埋下伏笔.  day -1  打印模板……打了200+页……事后证明大量的高中竞赛内容是没卵用的……  day 0  一大早打车去双流坐飞机,…

5510 Bazinga 题意:给出n个字符串,求满足条件的最大下标值或层数 条件:该字符串之前存在不是 它的子串 的字符串 求解si是不是sj的子串,可以用kmp算法之类的. strstr是黑科技,比手写的kmp快.if(strstr(s[i], s[j]) == NULL),则Si不是Sj的子串. 还有一个重要的剪枝:对于一个串,如果当前找到的串是它的母串,则下一次这个串不用遍历. #include <set> #include <queue> #include <cst…

HDU 6225 Little Boxes 题意 计算四个整数的和 解题思路 使用Java大整数 import java.math.BigInteger; import java.util.Scanner; /** * * @author reqaw */ public class Main { /** * @param args the command line arguments */ public static void main(String[] args) { Scanner in =…

Largest Point Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 3065    Accepted Submission(s): 1078 Problem Description Given the sequence A with n integers t1,t2,⋯,tn. Given the integral coeffic…

Best Solver Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)Total Submission(s): 1115    Accepted Submission(s): 645 Problem Description The so-called best problem solver can easily solve this problem, with his/her…

A B C D E F G H I J K L M O O O $\varnothing$ $\varnothing$   $\varnothing$ $\varnothing$ $\varnothing$ $\varnothing$      $\varnothing$ [A. Thickest Burger] 签到. [B. Relative atomic mass] 签到 [C. Recursive sequence] $$f[i] = f[i - 1] + 2 * f[i - 2] +…

链接在这:http://bak.vjudge.net/contest/132442#overview. A题,给出a,b和n,初始的集合中有a和b,每次都可以从集合中选择不同的两个,相加或者相减,得到一个新的数,如果在1~n内的话就放入集合中,并算一次操作,谁先不能操作(所有新数已经存在于集合内的话就不能进行操作)者输.问谁会赢. 可以得到的数字为k*gcd(a,b),那么只要算出在1~n的范围内存在多少个这样的数字,判断一下奇偶性即可. B题,给出n个字符串,问最大的满足条件的字符串的位置,条…

A.模拟 B.模拟 C(hdu5950):(矩阵快速幂) 题意:求f(n)=2f(n-2)+f(n-1)+n^4 分析:矩阵快速幂,(f(n),f(n-1),n^4,n^3,n^2,n,1) 注意:矩阵快速幂乘的时候初始矩阵是f[i][i]=1,而做加法的时候f[][]=0 E(hu5952):(dfs+剪枝) 题意:n<=100个点,m<=1000条边的图,求大小恰好是s<=10的完全图的个数,保证每个点的度数<=20 分析:这一看就是dfs,注意(1,2,3)和(3,2,1)等…

C.Recursive sequence 求ans(x),ans(1)=a,ans(2)=b,ans(n)=ans(n-2)*2+ans(n-1)+n^4 如果直接就去解...很难,毕竟不是那种可以直接化成矩阵的格式,我们也因为这个被卡很长时间 事实上可以把这道式子化成几个基本元素的格式,然后就容易组合了,比如ans(n-2)*2+ans(n-1)+(n-1)^4+4*(n-1)^3+6*(n-1)^2+4*(n-1)^1+1 包含了所有的基本组成形式,化绝对为相对,并且除了一个n-2其他都是n…