B. Bear and Forgotten Tree 3 time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output A tree is a connected undirected graph consisting of n vertices and n  -  1 edges. Vertices are numbered 1 through…

Bear and Forgotten Tree 3 time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A tree is a connected undirected graph consisting of n vertices and n  -  1 edges. Vertices are numbered 1 through…

E - Bear and Forgotten Tree 2 思路:先不考虑1这个点,求有多少个连通块,每个连通块里有多少个点能和1连,这样就能确定1的度数的上下界. 求连通块用链表维护. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define PLI pair<L…

C. Bear and Forgotten Tree 3 题目连接: http://www.codeforces.com/contest/658/problem/C Description A tree is a connected undirected graph consisting of n vertices and n  -  1 edges. Vertices are numbered 1 through n. Limak is a little polar bear and Rade…

E. Bear and Forgotten Tree 2 题目连接: http://www.codeforces.com/contest/653/problem/E Description A tree is a connected undirected graph consisting of n vertices and n  -  1 edges. Vertices are numbered 1 through n. Limak is a little polar bear. He once…

题目链接: C. Bear and Forgotten Tree 3 time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A tree is a connected undirected graph consisting of n vertices and n  -  1 edges. Vertices are numbered …

C. Bear and Forgotten Tree 3 time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A tree is a connected undirected graph consisting of n vertices and n  -  1 edges. Vertices are numbered 1 thro…

[题目链接] https://codeforces.com/problemset/problem/639/B [算法] 当d > n - 1或h > n - 1时 , 无解 当2h < d时无解 当d = 1 , n不为2时 , 无解 否则 , 我们先构造一条长度为h的链 , 然后 , 将一条(d - h)的链接到根上 , 再将剩余节点接到根上 时间复杂度 : O(N) [代码] #include<bits/stdc++.h> using namespace std; tem…

[链接] 我是链接,点我呀:) [题意] [题解] 首先,因为高度是h 所以肯定1下面有连续的h个点依次连成一条链.->用了h+1个点了 然后,考虑d这个约束. 会发现,形成d的这个路径,它一定是经过节点1比较好. 因为这条路径有两种可能-> 1.经过了1节点 2.没有经过1节点,那么肯定是1的某个子树里面,但是如果它的子树里再来一条长度为d的路径,肯定没有比经过1来的好,因为如果在1的子树里面的话有增加树的高度h的风险. 为了降低这个超过h的风险,那么我们还是优先让这个路径经过节点1. 然后…

题意:构造出一个 n 个结点,直径为 m,高度为 h 的树. 析:先构造高度,然后再构造直径,都全了,多余的边放到叶子上,注意直径为1的情况. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <io…