题目地址:http://poj.org/problem?id=1861 题意:输入点数n和边数n,m组边(点a,点b,a到b的权值).要求单条边权值的最大值最小,其他无所谓(所以多解:(.输出单条边最大值,边的数量,每条边(点a,点b). 思路:结构体记录节点x,y和权值d.写一个比较函数+sort使结构体按权值由小到大排序. 并查集:两个集合被选中的和没被选中的. kruskal:初始化每个节点独自一个集合,每次输入不在一个集合的就合并并记录,在一个集合的不管.输出记录数组最后一个节点的权值(…

题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=3371 思路: 这道题很明显是一道最小生成树的题目,有点意思的是,它事先已经让几个点联通了.正是因为它先联通了几个点,所以为了判断连通性 很容易想到用并查集+kruskal. 不过要注意 这题有一个坑点,就是边数很多 上限是25000,排序的话可能就超时了.而点数则比较少 上限是500,所以很多人选择用Prim做.但我个人觉得这样连通性不好判断.其实边数多没关系,我们只要去重就好啦,用邻接矩阵存下两点…

Constructing Roads Time Limit: 2000MS   Memory Limit: 65536K               Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A…

Conscription Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14661   Accepted: 5102 Description Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to b…

一.什么是并查集 在计算机科学中,并查集是一种树型的数据结构,用于处理一些不交集的合并及查询问题.有一个联合-查找算法(union-find algorithm)定义了两个用于次数据结构的操作: Find:确定元素属于哪一个子集.它可以被用来确定两个元素是否属于同一子集. Union:将两个子集合并成一个集合. 二.主要操作 初始化:把每个点所在的集合初始化为其自身. for(int i=1;i<=n;i++) f[i]=i; 查找:查找元素所在的集合,即根节点. int find(int x)…

并查集 先定义 int f[10100];//定义祖先 之后初始化 for(int i=1;i<=n;++i) f[i]=i; //初始化 下面为并查集操作 int find(int x)//int 类型 查找 { return f[x]==x?f[x]:f[x]=find(f[x]);//三目运算符查找 //如果f[x]==x 返回f[x] 否则返回f[x]=find(f[x]); } void unionn(int a,int b)//void 类型 连接 { a=find(a),b=fin…

Wireless Network Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 18066   Accepted: 7618 Description An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computer…

A very big corporation is developing its corporative network. In the beginning each of the N enterprises of the corporation, numerated from 1 to N, organized its own computing and telecommunication center. Soon, for amelioration of the services, the…

Corporative Network Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description   A very big corporation is developing its corporative network. In the beginning each of the N enterprises of th…

还是畅通project Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26860    Accepted Submission(s): 11985 Problem Description 某省调查乡村交通状况,得到的统计表中列出了随意两村庄间的距离.省政府"畅通project"的目标是使全省不论什么两个村庄间都能够实现公路交…