搜索资源列表
POJ
- PKU 1000-1050我ac的代码！绝对保证质量！-PKU 1000-1050 code I ac! Absolute guarantee of quality!
poj
- pku 3613 Cow Relays 题意：给一个无向图，求从起点s到终点e尽力n条边的路径最小值。 边的数量2<=t<=100(每条边可以重复遍历) 算法：利用dp求出 path[ l ] [ i ] [ k ]=MIN(path[ l ][ i ] [ k ] , path[ l ][ i ] [ j ] + path[ l ][ j ] [ k ]) l 表示的是2的指数，假设 i 到 k 要途径 2^l 条边，那么他可以被分为两段 途径2
POJ1000-1010
- ACM 相关的，POJ 1000-1010的全部解答，程序十分完备 部分还有文字说明-ACM-related, POJ 1000-1010 all the answers, procedures, and there is still a very complete text descr iption
123456ACM
- ACM POJ上的源码~~1000-3217一共719道题，这是网站http://acm.pku.edu.cn/JudgeOnline/，非常好的C++源码-ACM POJ on the source code ~ ~ 1000-3217 a total of 719 questions, this is the site http://acm.pku.edu.cn/JudgeOnline/, a very good C++ source code ~ ~
pojcodefor(1000-1099)
- poj从题号1000到1099道题目的源码，可能会有几道没有，但是近100道题目。每道题目至少有一道AC的代码，除了AC的还有很多效率不是很高的以及wa的代码，信息量很大，很好用。-poj from Question No. 1000-1099 subject source, may Jidao not, but nearly 100 questions. Each subject at least one AC code, in addition to the efficiency of AC
Wheres-Waldorf
- a+b 问题 poj 1000 a+b 问题-a+b problem
poj3941
- poj 3941 Hideyuki is allowed by his father Ujisato some 1000 yen bills every month for his pocket money. In the first day of every month, the number of bills is decided as follows. Ujisato prepares n pieces of m-sided dice and declares the cutback
poj3910
- poj 3910 nput The input file contains several test cases. Each test case starts with an integer n (0 < n < 1000), that stands for the cardinality of S. The next line contains the numbers of S: x1, x2, ..., xn. It is known that each xi is
Wheres-Waldorf
- a+b 问题 poj 1000 a+b 问题-a+b problem
ac1000-1199
- POJ上1000-1199，除1070,1109,1115,1138,1139,1162,1172,1181的所有代码-POJ s codes. 1000-1199, except:1070,1109,1115,1138,1139,1162,1172,1181