码迷,mamicode.com
首页 >  
搜索关键字:stand    ( 1367个结果
POJ 2528 Mayor's posters(线段树+离散化)
Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 46857   Accepted: 13601 Description The citizens of Bytetown, AB, could not stand that the ca...
分类:其他好文   时间:2015-04-08 13:17:40    阅读次数:135
D - Mayor's posters (线段树+离散化处理) POJ 2528
D - Mayor's posters Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2528 Description The citizens of Bytetown, AB, could not stand that the candid...
分类:其他好文   时间:2015-04-04 18:32:53    阅读次数:320
(简单) POJ 3169 Layout,差分约束+SPFA。
Description Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 #include #include #include #include #...
分类:其他好文   时间:2015-03-14 18:32:20    阅读次数:753
【java报错】Could not instantiate listener com.wbm.carriage.commons.log.LogbackConfigListener
这个错误以前出现过好几次,莫名其妙的出现,莫名其妙的解决掉。。。昨天好好的,今天又出现了,记下来2015-03-03 09:38:45.105:INFO:oejs.Server:jetty-8.1.14.v201310312015-03-03 09:38:45.783:WARN:oejw.Stand...
分类:编程语言   时间:2015-03-03 11:22:04    阅读次数:843
UVa 11174 (乘法逆元) Stand in a Line
题意:有n个人排队,要求每个人不能排在自己父亲的前面(如果有的话),求所有的排队方案数模1e9+7的值。分析:《训练指南》上分析得挺清楚的,把公式贴一下吧:设f(i)为以i为根节点的子树的排列方法,s(i)表示以i为根的子树的节点总数。f(i) = f(c1)f(c2)...f(ck)×(s(i)-...
分类:其他好文   时间:2015-03-02 23:40:15    阅读次数:196
UVa 11174 Stand in a Line
依旧是《训练指南》上的一道例题。书上讲的比较抽象,下面就把解法具体一下。因为涉及到父子关系,因此自然而然可以将n个节点构造成一棵树,最后将形成一个森林。接下来将使用递归的手法。设f(i)是以节点i为树根的子树,节点i有儿子c1,c2,c3....cj共j棵子树。s[i]为树根为i的子树包含的节点数。如果分别先给各个子树内部排序,那么毫无疑问, 共有f(c1)*f(c2)*f(c3)....*f(...
分类:其他好文   时间:2015-02-19 18:40:24    阅读次数:267
hdu5172---GTY's gay friends
Problem Description GTY has n gay friends. To manage them conveniently, every morning he ordered all his gay friends to stand in a line. Every gay friend has a characteristic value ai , to express how...
分类:其他好文   时间:2015-02-14 13:47:38    阅读次数:246
POJ3169(最短路+差分约束)
Layout Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7613   Accepted: 3658 Description Like everyone else, cows like to stand close to their friend...
分类:其他好文   时间:2015-02-11 18:35:35    阅读次数:177
HDU 5172 GTY's gay friends(线段树)
Problem Description GTY has n gay friends. To manage them conveniently, every morning he ordered all his gay friends to stand in a line. Every gay friend has a characteristic value ai , to expr...
分类:其他好文   时间:2015-02-08 18:14:38    阅读次数:114
POJ 3169 Layout (差分约束系统 + Bellman-ford算法)
Layout Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7613   Accepted: 3658 Description Like everyone else, cows like to stand close to their friends wh...
分类:编程语言   时间:2015-01-29 22:35:47    阅读次数:330
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!