Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11132    Accepted Submission(s): 4673 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the followin…

Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39047    Accepted Submission(s): 17279 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the followi…

Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51785    Accepted Submission(s): 23041 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the followi…

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description Given a string containing only 'A' - 'Z', we could encode it using the following method: 1. Each sub-string containing k same characters should be encoded to &quo…

pid=1020">Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25691    Accepted Submission(s): 11289 Problem Description Given a string containing only 'A' - 'Z', we could encode it usi…

HDU 1815, POJ 2749 Building roads pid=1815" target="_blank" style="">题目链接HDU 题目链接POJ 题意: 有n个牛棚, 还有两个中转站S1和S2, S1和S2用一条路连接起来. 为了使得随意牛棚两个都能够有道路联通,如今要让每一个牛棚都连接一条路到S1或者S2. 有a对牛棚互相有仇恨,所以不能让他们的路连接到同一个中转站. 还有b对牛棚互相喜欢,所以他们的路必须连到同一个中专站.…

HDU 1686 Oulipo / POJ 3461 Oulipo / SCU 2652 Oulipo (字符串匹配,KMP) Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book: Tout avait…

HDU 1816, POJ 2723 Get Luffy Out pid=1816" target="_blank" style="">题目链接 题意:N串钥匙.每串2把,仅仅能选一把.然后有n个大门,每一个门有两个锁,开了一个就能通过,问选一些钥匙,最多能通过多少个门 思路:二分通过个数.然后对于钥匙建边至少一个不选,门建边至少一个选,然后2-sat搞一下就可以. 一開始是按每串钥匙为1个结点,但是后面发现数据有可能一把钥匙,出如今不同串(真是不合…

题目 以前做过的一道题, 今天又加了一种方法 整理了一下..... 题意:给出一个字符串,问要将这个字符串变成回文串要添加最少几个字符. 方法一: 将该字符串与其反转求一次LCS,然后所求就是n减去 最长公共子串的长度. 额,,这个思路还是不是很好想. LCS: #include<iostream> #include<cstring> #include<cstdio> using namespace std; +; char s1[maxn], s2[maxn]; ][…

http://acm.hdu.edu.cn/showproblem.php?pid=1043 http://poj.org/problem?id=1077 Eight Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9173    Accepted Submission(s): 2473 Special Judge Problem D…