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…

Problem Description Given the sequence A with n integers t1,t2,⋯,tn. Given the integral coefficients a and b. The fact that select two elements ti and tj of A and i≠j to maximize the value of at2i+btj, becomes the largest point. Input An positive int…

Problem Description Elves are very peculiar creatures. As we all know, they can live for a very long time and their magical prowess are not something to be taken lightly. Also, they live on trees. However, there is something about them you may not kn…

http://acm.hdu.edu.cn/showproblem.php?pid=5441 Travel Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2061    Accepted Submission(s): 711 Problem Description Jack likes to travel around the wo…

http://acm.hdu.edu.cn/showproblem.php?pid=5444 Elven Postman Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 939    Accepted Submission(s): 520 Problem Description Elves are very peculiar crea…

Elven Postman Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 591    Accepted Submission(s): 329 Problem Description Elves are very peculiar creatures. As we all know, they can live for a very…

2017沈阳ICPC 10月20日 早上十点抵达沈阳,趁着老师还没到,跑去故宫游玩了一下,玩到一点多回到宾馆,顺便吃了群里大佬说很好吃的喜家德虾饺(真的好好吃),回到宾馆后身体有点不舒服了,头晕晕的,可能是吃太多热气的东西,也可能是去故宫玩的原因,wx也有一点(这是我作为一个队长的不称职了,非但没督促队友一起好好准备比赛,还带头玩乐,这也奠定了后面的失败).晚上老师带我们去吃包子,说实话不喜欢这个老师.吃完饭回去和天佑一起看了单纯型和后缀自动机,结果都没怎么看懂,太笨了,难受. 10月21日 早…

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…

Largest Point Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 536 Accepted Submission(s): 230 Problem Description Given the sequence A with n integers t1,t2,⋯,tn. Given the integral coefficients a…

题目;http://acm.hdu.edu.cn/showproblem.php?pid=5461 题意就是在数组中找出a*t[i]*t[i]+b*t[j]的最大值,特别注意的是这里i和i不能相等,想不到的是居然直接暴力排序都能过,而且这题的数据很水,好多有bug的代码都能过 但是我们要对自己要求严一点,尽量吧各种情况考虑到位,如果下面的代码还有不妥,欢迎指正 #include<cstdio> #include<cmath> #include<algorithm> us…

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…

队名:Unlimited Code Works(无尽编码)  队员:Wu.Wang.Zhou 先说一下队伍:Wu是大三学长:Wang高中noip省一:我最渣,去年来大学开始学的a+b,参加今年区域赛之前只学了大部分图论内容,以及一些数据结构.动态规划等内容,水平不及两个队友... ... 首先流水账式的记录一下比赛过程吧,最后再写这一年的感想.体会与将来的学习计划. 先从长春站说起... ... 长春站是我加入ACM以来参加的第一场ICPC,因此无比的激动!从杭州出发,乘了整整24小时的火车,终…

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…

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…

1001 Traversal 1002 Best Solver 1003 Minimum Cut 类似于POJ 3417的做法. 考虑每条新边对树边的覆盖次数. 每条树边被覆盖的次数其实就是断裂这条树边后还需断裂的新边数. 定义dp[i]为节点i向树根方向的边被新边覆盖次数.离线LCA后树DP. 答案为dp[2]~dp[n]中的最小值+1. # include <iostream> # include <cstdio> # include <cstring> # inc…

A prime number (or a prime) is a natural number greater than 11 that cannot be formed by multiplying two smaller natural numbers. Now lets define a number NN as the supreme number if and only if each number made up of an non-empty subsequence of all…

A. Boxes and Balls 二分找到最大的不超过$n$的$\frac{x(x+1)}{2}$形式的数即可. #include <bits/stdc++.h> using namespace std ; typedef long long LL ; void solve () { LL n ; scanf ( "%lld" , &n ) ; LL l = 1 , r = 2e9 ; while ( l < r ) { LL m = l + r + 1…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5458 Problem Description Given an undirected connected graph G with n nodes and m edges, with possibly repeated edges and/or loops. The stability of connectedness between node u and node v is defined by…

这题第一眼就想到暴力.. 枚举每一个ti,就能确定tj,tj一定是剩下数最大或最小的.为了求tj就要求出数列最大最小次大次小.时间复杂度O(n). 感觉暴力挺有趣的. #include<cstdio> #include<algorithm> using namespace std; __int64 arr[]; int main(){ int t; scanf("%d",&t); ; cse<=t; ++cse){ __int64 n,a,b; s…

时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 You must have seen the very famous movie series,"Mission Impossible", from 1 to 4. And "Mission Impossible 5" is now on screen in China. Tom Cruise is just learning programming through my MOOC cour…

题意: 给出一棵树,再给出每个节点上的值(一个char字符)这些值以一个字符串s1表示,然后给出一个s2字符串,问在这棵树上是否存在两个点,从一个点走到另一个点所经过的路径上的char字符组成的字符串正好等于s1.问是否存在这么两个点.如果存在,则输出“Find”,否则,输出“Important”. 题解: 使用dfs就可以解决,但是需要进行剪枝,否则就会tle. 剪枝的方法是这样的——假设节点1是根节点,然后我们先使用一次dfs记录每个节点到叶节点的最长的路径dis[x]. 然后开始搜索,每次…

题意: 给出一个n*m的地图,要求从左上角(0, 0)走到右下角(n-1, m-1). 地图中每个格子中有一个值.然后根据这些值求出一个最小值. 这个最小值要这么求—— 这是我们从起点走到终点的路径,其中N是地图的长,M是地图的宽,Ai表示路径中第i个点的值,Aavg表示路径中所有的点的值的平均值.要求这个式子的值最小. 我们可以将它转化为 好了,推到这里,我们需要的数学知识就结束了(实际上以我的数学知识也只能做到这里了……).然后dp就好了—— Dp[i][j][k],i表示第i行,j表示第j…

很坑的一道题,读了半天才读懂题,手忙脚乱的写完(套上模板+修改模板),然后RE到死…… 题意: 题面上告诉了我们这是一棵二叉树,然后告诉了我们它的先序遍历,然后,没了……没了! 反复读题,终于在偶然间注意到了这一句——"Not only that, when numbering the rooms, they always number the room number from the east-most position to the west." 它告诉我们,东边的点总是比西边的点…

Problem Description Princess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of some value v, and all of them will come at a different time. Because the lobby is not large enough, Alisha can only let a…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5489 题目大意: 一个N(N<=100000)个数的序列,要从中去掉相邻的L个数(去掉整个区间),使得剩余的数最长上升子序列(LIS)最长. 题目思路: [二分][最长上升子序列] 首先,假设去掉[i,i+m-1]这L个数,剩余的LIS长度为max(i左端最后一个不大于a[i+m]的LIS长度+a[i+m]开始到最后的LIS长度). 所以,我们从n到1逆向先求最长下降子序列的长度f[i],就可以知…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5493 题目大意: N个人,每个人有一个唯一的高度h,还有一个排名r,表示它前面或后面比它高的人的个数,求按身高字典序最小同时满足排名的身高排列. 题目思路: [线段树] 首先可以知道,一个人前面或后面有r个人比他高,那么他是第r+1高或第n-i-r+1高,i为这个人是第几高的. 所以先将人按照身高从小到大排序,接下来,把当前这个人放在第k=min(r+1,n-i-r+1)高的位置. 用线段树维护包…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5492 题目大意: 一个N*M的矩阵,一个人从(1,1)走到(N,M),每次只能向下或向右走.求(N+M-1)ΣN+M-1(Ai-Aavg)2最小.Aavg为平均值. (N,M<=30,矩阵里的元素0<=C<=30) 题目思路: [动态规划] 首先化简式子,得原式=(N+M-1)ΣN+M-1(Ai2)-(ΣN+M-1Ai)2 f[i][j][k]表示走到A[i][j]格子上,此时前i+j-1…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5491 题目大意: 一个数D(0<=D<231),求比D大的第一个满足:二进制下1个个数在[s1,s2]范围内.D已经满足[s1,s2]. 题目思路: [贪心][模拟] 首先将这个数转成二进制统计总共1的个数s,再求出末尾连续0和1的个数n0,n1. 如果最后一位是0: s=s2,那么为了保证s<s2且答案>D,先设ans=d+lowbit(d),此时满足了新的s<s2且答案&g…

1001 Monitor the Alpacas 1002 The Relationship in Club 1003 Difference of Clustering 两边离散化.暴力扫C就过了. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define maxn 1000010 bool vis1[m…