http://acm.hdu.edu.cn/showproblem.php? pid=4786 Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1733    Accepted Submission(s): 543 Problem Description Coach Pang is interested i…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=4786 题意:有N个节点(1 <= N <= 10^5),M条边(0 <= M <= 10^5).当中一部分边被染成了黑色,剩下的边是白色,问能不能建立一棵树,树中有斐波那契数个白色边. 思路:用克鲁斯卡尔建三次树,第一是用全部边建树.推断能否建成一棵树,第二次用黑边建树,最多能够用到x条黑边(不成环),n-1-x就是最少须要用的白边的数量,第三次用白边建树,最多能够用到y条白边.假设在[y…

http://acm.hdu.edu.cn/showproblem.php?pid=4786 Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 49    Accepted Submission(s): 26 Problem Description Coach Pang is interested in Fi…

Fibonacci Tree 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=4786 Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to do some research on Spanning Tree. So Coach Pang decides to solve the following problem: Consi…

Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to do some research on Spanning Tree. So Coach Pang decide…

Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 75    Accepted Submission(s): 38 Problem Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to do…

Fibonacci Tree Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2952    Accepted Submission(s): 947 Problem Description Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him t…

首先计算图的联通情况,如果图本身不联通一定不会出现生成树,输出"NO",之后清空,加白边,看最多能加多少条,清空,加黑边,看能加多少条,即可得白边的最大值与最小值,之后判断Fibonacci数是否在这两个之间,如果是输出yes,否则no. 然而,,然而,,我看的题解有问题!!!!!调了俩小时愣是没找出错误来,,然后把题解交了发现过不了,,,,真是够了.,.,., 第二天上午终于A了,,满分程序: #include <cstdio> #include <cstring&…

题意: N个点,M条边.每条边连接两个点u,v,且有一个权值c,c非零即一. 问能否将N个点形成一个生成树,并且这棵树的边权值和是一个fibonacii数. (fibonacii数=1,2,3,5,8 .... ) 思路: 若可以生成一棵树.则有最小生成树和最大生成树.假设已经生成了最小MST  P 和最大MST  Q. 将P更换一条边可以得到另一棵生成树,边权和不是和P相等就是比P的边权和大1.(因为边值非零即一).同理搞下去....一定可以得到Q. 所以P的边权和到Q的边权和之间的所有值都能…

一个由n个顶点m条边(可能有重边)构成的无向图(可能不连通),每条边的权值不是0就是1. 给出n.m和每条边的权值,问是否存在生成树,其边权值和为fibonacci数集合{1,2,3,5,8...}中的一个. 求最小生成树和最大生成树,得到生成树边权和的下界left和上界right.这道题由于每条边权值不是0就是1,因此生成树边权和可以覆盖到[left,right]的每一个数.那么求得[left,right],看是否有fibonacci数在区间内就可以了. #include <cstdio>…