资源的copying行为决定对象的copying行为. 抑制copying行为,使用引用计数.…

题目链接: 传送门 Copying Books Time Limit: 3000MS     Memory Limit: 32768 KB Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had b…

复制RAII对象必须一并复制它所管理的资源,所以资源的copying行为决定RAII对象的copying行为. 普遍而常见的RAII class copying行为是:抑制copying(使用私有继承Uncopyable).施行引用计数法(reference counting)(即std::tr1::shared_ptr,可以自己指定删除器).不过其他行为也都可能被实现.…

Copying Data Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Description We often have to copy large volumes of information. Such operation can take up many computer resources. Therefore, in this problem you are advised to come…

Copying  Books 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=85904#problem/B 题目: Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so calledscri…

This is a time saving tip for application designer. If you are creating a new record definition and you want all or most of the fields from an existing record, you can simply open the existing record, select the fields you want and drag them over to…

题目链接: 题目 Copying Books Time limit: 3.000 seconds 问题描述 Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after s…

http://poj.org/problem?id=1505 Copying Books Time Limit: 3000MS   Memory Limit: 10000K Total Submissions: 7053   Accepted: 2200 Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be r…

第一节 <背景> 条款13中讲到“资源取得的时机便是初始化时机”并由此引出“以对象管理资源”的概念.通常情况下使用std中的auto_ptr(智能指针)和tr1::shared_ptr(引数智能指针)作为管理资源的对象.事实上,这种管理方法十分有效.但是,auto_ptr和tr1::shared_ptr只能管理基于堆(heap-based)的资源,而非heap-based的资源却往往不适合.因此,有的时候你需要建立自己的资源管理类.本文介绍的内容是在你建立自己的资源管理类时应该注意的事项. 第…

Copying Linked Lists with Random Pointers 两个方法:方法一: 1.不考虑随机指针的情况下复制链表: 2.在复制过程中建立一个以原链表节点地址为key,相应的复制链表节点地址为value的hash: 3.遍历原链表和新链表,得到原链表随机指针值,并复制给新链表. 方法二: 1.忽略随机指针值复制链表第n个节点并插入到第n个节点与第n+1个节点之间,以此为方式修改链表直到链表尾. 2.利用链表节点next指针指向节点的的拷贝这一已知,使用如下语句: srcC…