思路: 拓扑排序.贪心. 实现: #include <bits/stdc++.h> using namespace std; vector<]; int n1, n2; inline void help(queue<int>& a, queue<int>& b, int x) { if (x > n1) b.push(x); else a.push(x); } bool same(int x, int y) { && x &l…

点我看题目 题意 :其实题意我也说不清楚,因为比赛的时候我盯着看了1个小时也没看懂....就是两个磁盘,第一个有n1的安装包,编号为1~n1,第二个有n2个安装包,编号为n1~n2.给你d对关系,(x,y)代表着安装x之前要先安装y.然后让你往里插入这两个磁盘,因为一次只能插一次,所以要满足他给的条件的时候要频繁的来回换.还有,要注意的一点是,无论先往里边插第一个还是第2个,第一次插的时候算一次,最后一次拔的时候算一次. 思路 :其实我真不知道这是拓扑排序,,,,,后来才知道的..... //#…

http://poj.org/problem?id=1778 题意:有两个DVD,第一个DVD上有编号为1~n1的安装包,第二个DVD上有编号为n1+1~n1+n2的安装包,给出m组关系(a,b) 表示安装a之前必须先安装b.由于安装时每次只能插入一个DVD,问安装完所有的安装包,这两个DVD至少要交换插入多少次.ps:第一次插入算一次,最后一次拔出算一次. 思路:两次拓扑排序,以先插入第一个DVD,进行拓扑排序,求出交换次数:以先插入第二个DVD,进行拓扑排序,求出交换次数.最后输出这两种交换…

杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029. 1032.1037.1040.1048.1056.1058.1061.1070.1076.1089.1090.1091.1092.1093. 1094.1095.1096.1097.1098.1106.1108.1157.116…

Statistical approaches to randomised controlled trial analysis The statistical approach used in the design and analysis of the vast majority of clinical studies is often referred to as classical or frequentist. Conclusions are made on the results of…

WARN:30 20:55:45,340ms- [HqlSqlWalker]1009行-[DEPRECATION] Encountered positional parameter near line 1, column 33 in HQL: [FROM com.hs.model.IpModel WHERE ? BETWEEN ip_start_num AND ip_end_num].  Positional parameter are considered deprecated; use na…

异常处理被认为存在缺陷 Do, or do not. There is no try. - Yoda, The Empire Strikes Back (George Lucas) by Jason Robert Carey Patterson, Nov 2005 Recent programming languages such as Java, Python and Ruby have chosen to use exception handling as their primary met…

1, // tcpmux 7, // echo 9, // discard 11, // systat 13, // daytime 15, // netstat 17, // qotd 19, // chargen 20, // ftp data 21, // ftp access 22, // ssh 23, // telnet 25, // smtp 37, // time 42, // name 43, // nicname 53, // domain 77, // priv-rjs 7…

随机增量算法(a randomized incremental algorithm) #define sqr(x) ((x)*(x)) #define EPS 1e-4 struct P{ double x, y; P(double x, double y):x(x),y(y){} P(P &a, P &b):x(b.x-a.x),y(b.y-a.y){} P(){} P mid(P &a){ , (a.y+y)/); } double cross(P &a){ retur…