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>…

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…

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…

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…

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…

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…

cf 443 D. Teams Formation(细节模拟题) 题意: 给出一个长为\(n\)的序列,重复\(m\)次形成一个新的序列,动态消除所有k个连续相同的数字,问最后会剩下多少个数(题目保证消除的顺序对答案不会造成影响) \(n <= 10^{5} ,m <= 10^{9} ,k <= 10^{9},a_i <= 10^{5}\) 思路: 直接上题解好了 首先要用栈预处理序列本身,然后考虑处理后的序列 比较头和尾是否能够消除,细节需要处理好. #include<bi…

题目传送门 /* 构造水题:对于0的多个位数的NO,对于位数太大的在后面补0,在9×k的范围内的平均的原则 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MAXN]; int main(void) //Codeforces Round #…

题目 构造一组 $x, y, z$,使得对于给定的 $n$,满足 $\frac{1}{x}  + \frac{1}{y} + \frac{1}{z} =  \frac{2}{n}$. 分析: 样例二已经暴露了此题的本质. 显然 $n, (n+1), n(n+1)$ 为一组合法解.特殊地,当 $n=1$ 时,无解,因为此时 $n+1$ 与 $n(n+1)$ 相等(也可以证明没有其他形式的解). #include<bits/stdc++.h> using namespace std; int n;…

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[…