题目:给出N个只有左右括号字符串 ,这N个字符串的排列顺序是任意的 , 问按最优的排序后 , 得到最多匹配的括号个数 分析: 我们很容易的想到 字符串)()()(( , 这样的字符串可以精简为)(( 因为无论如何的排序 ,对于字符串可以匹配的括号是不会变的 : 那么问题就可以简化为对与 **)(**    )     (   这几种类型的字符串的排序情况 : 我们也很自然而然的想到了贪心 ,那问题来了 ,我们该如何贪心呢?先从小问题出发 , 有A 与 B两串 , 在自然的可以想到 排序的情况肯定…

Balanced Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6311    Accepted Submission(s): 1648 Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of…

大意: 记$f(t)$表示字符串$t$的最长括号匹配子序列, 给定n个括号序列, 求它们重排后的最大f(t). 首先可以注意到一个括号序列中已经匹配的可以直接消去, 一定不会影响最优解. 那么这样最终就为n个类似于))))((的括号序列, 然后贪心排序即可 #include <iostream> #include <algorithm> #include <math.h> #include <cstdio> #include <set> #inc…

题意:题意一开始不是很明白...就是他给你n个串,让你重新排列组合这n个串(每个串内部顺序不变),使得匹配的括号长度最大.注意,题目要求not necessary continuous,括号匹配不需要连续. 思路:我们先把每个串里面能组合的全部抵消,比如)((()抵消完为)((.我们能知道,这样操作完只会有4种情况留下:))))),((((((((,)((((((,))))))))(.然后排序,排序的时候为了后面匹配能最大化利用所有括号,我们需要把左括号尽可能多的放在左边,右括号尽可能多的放在右…

#include <stdio.h> #include <iostream> #include <cstdlib> #include <cmath> #include <string> #include <cstring> #include <algorithm> #include <stack> #include <queue> #include <set> #include <…

HDU6299.Balanced Sequence 这个题就是将括号处理一下,先把串里能匹配上的先计数去掉,然后统计左半边括号的前缀和以及右半边括号的前缀和,然后结构体排序,然后遍历一遍,贪心策略走一遍就可以了. 但是我写的时候排序写挫了,左(括号)多右(括号)少的和左少右多的,肯定左多的在前面,左少右多和左多右少的,肯定左多的在前面,左少右多和左少右多,谁的左边的多谁在前面,以及其他情况都按右少的前排,按这四种情况考虑就可以完美贪心了.只是可惜自己智障,其他就没什么了. 代码: //1002-…

题目传送门 题目大意:给出n个字符串,定义了平衡字符串,问这些字符串组合之后,最长的平衡字符子序列的长度. 思路: 首先肯定要把所有字符串先处理成全是不合法的,记录右括号的数量为a,左括号的数量为b,考虑两个字符串,这两个如果min(a1,b2)小于min(a2,b1)时,第一个字符串是不是应该排在前面呢?如果两个相同的话,当然应该吧右括号比较多的放在前面,左括号比较多的放在后面. #include<iostream> #include<cstdio> #include<cm…

Description Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes…

题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=6047 题目: Maximum Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 90    Accepted Submission(s): 44 Problem Description Steph is extremely o…

题意:给定一个序列,让你构造出一个序列,满足条件,且最大.条件是 选取一个ai <= max{a[b[j], j]-j} 析:贪心,贪心策略就是先尽量产生大的,所以就是对于B序列尽量从头开始,由于数据比较大,采用桶排序,然后维护一个单调队列,使得最头上最大. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #i…

Balanced Sequence Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of this type is said to be balanced:…

Balanced Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6207    Accepted Submission(s): 1616 Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of…

Balanced Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1320    Accepted Submission(s): 316 Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of t…

HDU 4442 Physical Examination(贪心) 题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=4442 Description WANGPENG is a freshman. He is requested to have a physical examination when entering the university. Now WANGPENG arrives at the hospital. Er-.. Th…

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HDU 1711 Number Sequence(数列) Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) [Description] [题目描述] Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N…

HDU 1005 Number Sequence(数列) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) [Description] [题目描述] A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, an…

HDU5014Number Sequence(贪心) 题目链接 题目大意: 给出n,然后给出一个数字串,长度为n + 1, 范围在[0, n - 1].然后要求你找出另外一个序列B,满足上述的要求,而且使得t = A0^B0 + Ai + 1 ^ Bi + 1 + ... + An ^ Bn 最大. 解题思路: 对于一个数字进行异或,要求结果最大的话,那么取这个数字的二进制互补数字是最好的情况,而且能够发现每次找到一个数字和相应的互补的数字都会是一段区间.就这样一段一段区间的去寻找每一个点相应的…

HDU 5860 Death Sequence(递推) 题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=5860 Description You may heard of the Joseph Problem, the story comes from a Jewish historian living in 1st century. He and his 40 comrade soldiers were trapped in a cave…

HDU 1560 DNA sequence(DNA序列) Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)   Problem Description - 题目描述 The twenty-first century is a biology-technology developing century. We know that a gene is made of DNA. Th…

HDU 1005 Number Sequence(数论) Problem Description: A number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multipl…

HDU 1711 Number Sequence (字符串匹配,KMP算法) Description Given two sequences of numbers : a1, a2, ...... , aN, and b1, b2, ...... , bM (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make aK = b1, aK+1 = b2, ...... , aK+M…

[题解]Cut the Sequence(贪心区间覆盖) POJ - 3017 题意: 给定一大堆线段,问用这些线段覆盖一个连续区间1-x的最小使用线段的数量. 题解 考虑一个这样的贪心: 先按照左端点排序,若左端点一样则谁长谁在前.现在判无解就方便了,记录一下前缀max即可.然后现在要最小化选择. 记录一个最右端点\(R\),一个暴力的办法是暴力循环判断所有线段是否满足条件,这样显然超时,你决定优化一下常数,所以你记录一下从哪个线段开始才\(l_i \ge R\).你以为你是常数优化,其实你复…

/* HDU 6078 - Wavel Sequence [ DP ] | 2017 Multi-University Training Contest 4 题意: 给定 a[N], b[M] 要求满足 a[f(1)]<a[f(2)]>a[f(3)]<a[f(4)]>a[f(5)]<a[f(6)]... b[g(i)] == a[f(i)] f(i) < f(i+1), g(i) < g(i+1) 的子序列 的数目 分析: dp[i][j][0] 表示 以a[i]…

/* HDU 6047 - Maximum Sequence [ 单调队列 ] 题意: 起初给出n个元素的数列 A[N], B[N] 对于 A[]的第N+K个元素,从B[N]中找出一个元素B[i],在 A[] 中找到一个数字A[p]满足 B[i] <= p <= N+K-1 令 A[N+K] = A[p]-p,直到A[]的长度等于2N 问 A[N+1] + A[N+2] + ... + A[N<<1] 最大是多少 分析: 将A[]中元素全部减去其下标 将B[]排序,可分析一定是从小…

HDU - 1005 Number Sequence Problem Description A number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.Given A, B, and n, you are to calculate the value of f(n). Input The input consists of multiple test…

题目说要n个字符串串内随意组合以后将这些串放在一起,然后求最长的括号匹配的长度,并不要求是连续的 因为不需要是连续的,所以可以先把已经匹配好的括号加入到答案里面去,先把这些删掉,以为并不影响结果,然后最后就只剩下))))((((这些的序列,然后接下来对剩下的考虑 因为剩下的括号最多只有四种情况就是 1.只有(   (((((((( 2.只有)   ))))))) 3.(比)多  (((((()) 4.)比(多   ))))))((( 然后我们要做的就是把他们排序,既然要得到尽可能多的匹配,那就贪…

Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of this type is said to be balanced: + if it is the empty string+ if A and B are balanced, AB is balanced,+ if A is balanced, (A) is balanced. Chiaki can reorder…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 我们贪心地把每一个括号序列能匹配都按照栈的规则都匹配出来. (直接递增匹配对数*2就可以了 最后栈里面就只剩下类似))))(((((((这样的形式了. 现在就相当于有很多个这种字符串了. 让你把它们拼接在一起. 可以用sort贪心一下. 优先把(多的序列放在前面一点. 具体的比较函数这么写 里注释的括号用于参考 (实在不知道原理,就一种一种猜吧.... sort完之后再模拟一下括号匹配就好 bool operator < (co…

Divide the Sequence 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5783 Description Alice has a sequence A, She wants to split A into as much as possible continuous subsequences, satisfying that for each subsequence, every its prefix sum is not smal…