题意:给出一组含m个点的无向图,再给出n个点,这n个点分别以一条边连接到这个无向图中的某个点.对于每个询问,求出3点连通的最小代价.有可能3个点是不能互通的.如图,最小代价就是红色的边的权之和. 思路:先对m个点的无向图进行求两两之间最短路径,用floyd.接下来对于每个询问,穷举m个点,求3个点分别到该点的距离之和,求最小即可. //#include <bits/stdc++.h> #include <iostream> #include <cstdio> #incl…

题目连接:problemId=542" target="_blank">ZOJ 1542 POJ 1861 Network 网络 Network Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge Andrew is working as system administrator and is planning to establish a new network in his com…

Escape Time II Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2966 Description In last winter, there was a big snow storm in South China. The electric system was damaged seriously. Lots of p…

QS Network Sunny Cup 2003 - Preliminary Round April 20th, 12:00 - 17:00 Problem E: QS Network In the planet w-503 of galaxy cgb, there is a kind of intelligent creature named QS. QScommunicate with each other via networks. If two QS want to get conne…

主题链接:problemId=203" target="_blank">ZOJ 1203 Swordfish 旗鱼 Swordfish Time Limit: 2 Seconds      Memory Limit: 65536 KB There exists a world within our world A world beneath what we call cyberspace. A world protected by firewalls, password…

Connect them ZOJ - 3204 You have n computers numbered from 1 to n and you want to connect them to make a small local area network (LAN). All connections are two-way (that is connecting computers i and j is the same as connecting computers j and i). T…

QS Network Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 1586 Appoint description:  System Crawler  (2015-05-31) Description Sunny Cup 2003 - Preliminary Round April 20th, 12:00 - 17:00 Problem…

Sunny Cup 2003 - Preliminary Round April 20th, 12:00 - 17:00 Problem E: QS Network In the planet w-503 of galaxy cgb, there is a kind of intelligent creature named QS. QScommunicate with each other via networks. If two QS want to get connected, they…

注意排序即可 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> using namespace std; #define for0n for(i=0;i<n;i++) #define for1n for(i=1;i<=n;i++) #define for0m fo…

题意:裸最小生成树,主要是要按照字典序. 思路:模板 prim: #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define INF 0x7fffffff #define MAXN 128 bool vis[MAXN]; int lowu[MAXN];//记录起始边(已加入集合中的边) int lowc[MAX…

Traveler Nobita Time Limit: 2 Seconds      Memory Limit: 65536 KB One day, Nobita used a time machine and went back to 1000 AD. He found that there are N cities in the kingdom he lived. The cities are numbered from 0 toN - 1. Before 1000 AD., there a…

链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1586 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82831#problem/E In the planet w-503 of galaxy cgb, there is a kind of intelligent creature named QS. QScommunicate with each o…

Build The Electric System Time Limit: 2 Seconds      Memory Limit: 65536 KB In last winter, there was a big snow storm in South China. The electric system was damaged seriously. Lots of power lines were broken and lots of villages lost contact with t…

Connect them Time Limit: 1 Second      Memory Limit: 32768 KB You have n computers numbered from 1 to n and you want to connect them to make a small local area network (LAN). All connections are two-way (that is connecting computers iand j is the sam…

题目链接: PKU:http://poj.org/problem?id=1861 ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=542 Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the c…

Traveler Nobita Time Limit: 2 Seconds      Memory Limit: 65536 KB One day, Nobita used a time machine and went back to 1000 AD. He found that there are N cities in the kingdom he lived. The cities are numbered from 0 to N - 1. Before 1000 AD., there…

学习最小生成树已经有一段时间了 做一些比较简单的题还算得心应手..花了三天的时间做完了kuangbin的专题 写一个题解出来记录一下(虽然几乎都是模板题) 做完的感想:有很多地方都要注意 n == 1 注意double 的精度问题 poj 1251 模板题 大写字母减去'A'+1即是它的编号 #include<stdio.h> #include<string.h> #include<algorithm> #include<map> #include<m…

转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…

POJ 1502 MPI Maelstrom / UVA 432 MPI Maelstrom / SCU 1068 MPI Maelstrom / UVALive 5398 MPI Maelstrom /ZOJ 1291 MPI Maelstrom (最短路径) Description BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shar…

首先,贴上一个很好的讲解贴: http://www.wutianqi.com/?p=3012 HDOJ 1233 还是畅通工程 http://acm.hdu.edu.cn/showproblem.php?pid=1233 裸的Prim... #include<cstdio> #define MAXN 105 #define INF 0x3f3f3f3f int map[MAXN][MAXN]; int dist[MAXN]; int vis[MAXN]; int n,a,b,x,ans,tot…

A & M - Jungle Roads HDU - 1301 题意:字母之间的路,求最小生成树 题解:处理好建边以后就是一个Prime #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<sstream> #include<cmath> #include<stack> #include<cstd…

A是水题,此处略去题解 B - PreSuffix ZOJ - 3995 (fail树+LCA) 给定多个字符串,每次询问查询两个字符串的一个后缀,该后缀必须是所有字符串中某个字符串的前缀,问该后缀最长时,是多少个字符串的前缀. 思路:对所有串构造ac自动机,根据fail指针的性质,a节点的fail指针指向b时,b一定是a的某个后缀.所以每次询问对两个字符串对应的节点在fail树上求一下LCA,插入时经过了LCA节点的字符串的个数便是答案. #include<bits/stdc++.h> us…

Tangled in Cables Time Limit: 2 Seconds      Memory Limit: 65536 KB You are the owner of SmallCableCo and have purchased the franchise rights for a small town. Unfortunately, you lack enough funds to start your business properly and are relying on pa…

以此图为例: package com.datastruct; import java.util.Scanner; public class TestKruskal { private static class Edge{ public Edge(int begin,int end,int weight){ this.begin = begin; this.end = end; this.weight = weight; } int begin; int end; int weight; publ…

ENode框架Conference案例分析系列之 - 业务简介 ENode框架Conference案例分析系列之 - 上下文划分和领域建模 ENode框架Conference案例分析系列之 - 架构设计 ENode框架Conference案例分析系列之 - Quick Start ENode框架Conference案例分析系列之 - 复杂情况的读库更新设计 ENode框架Conference案例分析系列之 - 订单处理减库存的设计 ENode框架Conference案例分析系列之 - ENode…

最小生成树计数 (1s 128M) award [问题描述] 现在给出了一个简单无向加权图.你不满足于求出这个图的最小生成树,而希望知道这个图中有多少个不同的最小生成树.(如果两颗最小生成树中至少有一条边不同,则这两个最小生成树就是不同的).由于不同的最小生成树可能很多,所以你只需要输出方案数对31011的模就可以了. [输入格式] 第一行包含两个数,n和m,其中1<=n<=100; 1<=m<=1000; 表示该无向图的节点数和边数.每个节点用1~n的整数编号.接下来的m行,每行…

第十三届浙江省大学生程序设计竞赛 I 题, 一道模拟题. ZOJ  3944http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=3944 In a BG (dinner gathering) for ZJU ICPC team, the coaches wanted to count the number of people present at the BG. They did that by having the waitre…

poj   1251  Jungle Roads  (最小生成树) Link: http://poj.org/problem?id=1251 Jungle Roads Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23507   Accepted: 11012 Description The Head Elder of the tropical island of Lagrishan has a problem. A b…

http://www.lydsy.com/JudgeOnline/problem.php?id=1016 统计每一个边权在最小生成树中使用的次数,这个次数在任何一个最小生成树中都是固定的(归纳证明). 在同一个边权上对所有边权为这个的边暴力统计(可以用矩阵树定理),然后用并查集把这个边权的所有边贡献的连通性都加上,再统计下一个边权. 最后把答案乘起来. #include<cstdio> #include<cstring> #include<algorithm> usin…

Prim算法 1.概览 普里姆算法(Prim算法),图论中的一种算法,可在加权连通图里搜索最小生成树.意即由此算法搜索到的边子集所构成的树中,不但包括了连通图里的所有顶点(英语:Vertex (graph theory)),且其所有边的权值之和亦为最小.该算法于1930年由捷克数学家沃伊捷赫·亚尔尼克(英语:Vojtěch Jarník)发现:并在1957年由美国计算机科学家罗伯特·普里姆(英语:Robert C. Prim)独立发现:1959年,艾兹格·迪科斯彻再次发现了该算法.因此,在某些场…