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题意：求一个无向图的最<em>小生</em>成树与次<em>小生</em>成树的边权和是否相等题解：首先有一个性质，就是最<em>小生</em>成树上的任意两点的距离就是其在原图中的最短路，严格的证明我
https://www.u72.net/daima/5de5.html - 2024-09-06 07:35:22 - 代码库DescriptionDuring the Warring States Period of ancient China(476 BC to 221 BC), there were seven kingdoms in China -- they were Qi, Chu, Yan
https://www.u72.net/daima/s164.html - 2024-07-13 07:44:31 - 代码库题目连接：ZOJ 1542 POJ 1861 Network 网络NetworkTime Limit: 2 Seconds Memory Limit: 65536 KB Special JudgeAndrew is working as system
https://www.u72.net/daima/5u60.html - 2024-07-23 07:29:10 - 代码库http://www.lydsy.com/JudgeOnline/problem.php?id=1016想也想不到QAQ首先想不到的是：题目有说，具有相同权值的边不会超过10条。其次：老是去想组合计数怎
https://www.u72.net/daima/nze9m.html - 2024-08-02 07:34:10 - 代码库Qin Shi Huang‘s National Road SystemTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s)
https://www.u72.net/daima/nzmhu.html - 2024-08-02 07:44:38 - 代码库发现，若使方差最小，则使&Sigma;(wi-平均数)最小即可。因为权值的范围很小，所以我们可以枚举这个平均数，每次把边权赋成(wi-平均数)2，做kruscal。但是，我们怎
https://www.u72.net/daima/nk3u3.html - 2024-08-04 07:03:35 - 代码库链接：http://poj.org/problem?id=1789题意：卡车公司有悠久的历史，它的每一种卡车都有一个唯一的字符串来表示，长度为7，它的全部卡车（除了第一辆）都是由曾经的
https://www.u72.net/daima/nar15.html - 2024-07-30 13:39:35 - 代码库1083: [SCOI2005]繁忙的都市Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 2925 Solved: 1927[Submit][Status][Discuss]Description 城市C
https://www.u72.net/daima/ns940.html - 2024-10-20 02:55:39 - 代码库id=3669lct维护最<em>小生</em>成树 裸题 最小的边一定在最<em>小生</em>成树上 如果我们能用其他边调整 那么我们从能调
https://www.u72.net/daima/emaz.html - 2024-09-16 03:46:09 - 代码库Qin Shi Huang‘s National Road SystemTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s):
https://www.u72.net/daima/xcfr.html - 2024-07-17 00:38:45 - 代码库http://poj.org/problem?id=3026http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1248Borg MazeT
https://www.u72.net/daima/xfne.html - 2024-07-16 23:29:10 - 代码库题目链接：啊哈哈，点我点我题意：北极的某区域共有n座村庄( 1 ≤ n ≤ 500 )，每座村庄的坐标用一对整数(x, y)表示，其中 0 ≤ x, y ≤ 10000。为了加强联系，
https://www.u72.net/daima/xk2a.html - 2024-07-16 21:17:02 - 代码库Sunny Cup 2003 - Preliminary RoundApril 20th, 12:00 - 17:00Problem E: QS NetworkIn the planet w-503 of galaxy cgb, there is a kind of intel
https://www.u72.net/daima/1mnb.html - 2024-07-19 16:58:06 - 代码库题目链接：ZOJ 1406 POJ 1251 Jungle Roads 丛林中的道路Jungle RoadsTime Limit: 2 Seconds Memory Limit: 65536 KBThe Head Elder of the tropic
https://www.u72.net/daima/5u6u.html - 2024-07-23 07:28:54 - 代码库题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1162【题目大意】给你n个点的坐标，让你找到联通n个点的一种方法。保证联通的线路最短，典型
https://www.u72.net/daima/naf72.html - 2024-09-18 11:03:27 - 代码库Qin Shi Huang‘s National Road SystemTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 【Problem Descripti
https://www.u72.net/daima/na9c3.html - 2024-07-31 03:57:02 - 代码库二分(分块)枚举 边权上限。用kruscal判可行性。#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int u[200
https://www.u72.net/daima/nr2ff.html - 2024-08-09 13:51:30 - 代码库题目链接：Sicily 1090思路：简单的最<em>小生</em>成树问题，这里用prim算法即可。用visited数组记录每个结点是否已经被访问，即是否已经在最<em>小生</em>成树中。
https://www.u72.net/daima/0h17.html - 2024-07-17 20:05:17 - 代码库今天的主要内容为最<em>小生</em>成树、判负环和差分约束系统 苗条的最<em>小生</em>成树 poj3522本题排序完枚举最小边，Kruskal跑n遍即可。
https://www.u72.net/daima/nss7e.html - 2024-10-17 10:05:39 - 代码库简单最<em>小生</em>成树，畅通project。这三道题目都是练习最<em>小生</em>成树的。注意一下推断是否有通路时，kruskal能够推断每一个点的祖先是否同样。
https://www.u72.net/daima/nbd5w.html - 2024-10-02 22:49:02 - 代码库