题意:n点,m条边.m条边里面标记为1的最小生成树的边,0为非最小生成树的边.给了每条边的权,如果能构成一个最小生成树则输出图,否则-1. 思路:先按权值小,为生成数边的顺序排序.(根据kruskal)再添加每条0边.这里假定(1,3),(2,4)构成环. #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #incl…

B. Lazy Student time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Student Vladislav came to his programming exam completely unprepared as usual. He got a question about some strange algorith…

构造.对边的权值排序,权值一样的话,在MST中的边排到前面,否则权值小的排在前面. 然后边一条一条扫过去,如果是1 ,那么连一个点到集合中,如果是0,集合内的边相连. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; +; struct Edge { int u,v; int w; int info; int id; } e[…

Lazy Student time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Student Vladislav came to his programming exam completely unprepared as usual. He got a question about some strange algorithm o…

D. Lazy Student   Student Vladislav came to his programming exam completely unprepared as usual. He got a question about some strange algorithm on a graph — something that will definitely never be useful in real life. He asked a girl sitting next to…

D. Lazy Student Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/606/problem/D Description Student Vladislav came to his programming exam completely unprepared as usual. He got a question about some strange algorithm on a gr…

题目链接: http://codeforces.com/contest/606/problem/D D. Lazy Student time limit per test2 secondsmemory limit per test256 megabytes 问题描述 Student Vladislav came to his programming exam completely unprepared as usual. He got a question about some strange…

题意: 一个n个节点的图,有m条边,已知这个图的一个mst 现在如果我们知道这个图的m条边,和知道mst的n-1条边是哪些,问能不能构造出一个满足条件的图 思路:排序+构造 数组deg[i]表示节点i此时还可以1-i-1中的多少条边相连 由于:deg[i]=i-1时,最低可以和1连接,=i-2时,最低可以和2连接 所以如果我们知道i和deg[i]的话,我们就知道这一次i要和哪一个节点连接, 就可以我构造出来了. #include <cstdio> #include <cstring>…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Student Vladislav came to his programming exam completely unprepared as usual. He got a question about some strange algorithm on a graph - somet…

在.NET4.0中,可以使用Lazy<T> 来实现对象的延迟初始化,从而优化系统的性能.延迟初始化就是将对象的初始化延迟到第一次使用该对象时.延迟初始化是我们在写程序时经常会遇到的情形,例如创建某一对象时需要花费很大的开销,而这一对象在系统的运行过程中不一定会用到,这时就可以使用延迟初始化,在第一次使用该对象时再对其进行初始化,如果没有用到则不需要进行初始化,这样的话,使用延迟初始化就提高程序的效率,从而使程序占用更少的内存. 下面我们来看代码,新建一个控制台程序,首先创建一个Student类…