ICPC (Isles of Coral Park City) consist of several beautiful islands. The citizens requested construction of bridges between islands to resolve inconveniences of using boats between islands, and they demand that all the islands should be reachable from any other islands via one or more bridges. The city mayor selected a number of pairs of islands, and ordered a building company to estimate the costs to build bridges between the pairs. With this estimate, the mayor has to decide the set of bridges to build, minimizing the total construction cost. However, it is difficult for him to select the most cost-efficient set of bridges among those connecting all the islands. For example, three sets of bridges connect all the islands for the Sample Input 1. The bridges in each set are expressed by bold edges in Figure F.1.
Figure F.1. Three sets of bridges connecting all the islands for Sample Input 1
As the first step, he decided to build only those bridges which are contained in all the sets of bridges to connect all the islands and minimize the cost. We refer to such bridges as no alternative bridges. In Figure F.2, no alternative bridges are drawn as thick edges for the Sample Input 1, 2 and 3.
Figure F.2. No alternative bridges for Sample Input 1, 2 and 3
Write a program that advises the mayor which bridges are no alternative bridges for the given input.
Input
The input file contains several test cases, each of them has the following format.
N M S1 D1 C1 . . . SM DM CM
The first line contains two positive integers N and M. N represents the number of islands and each island is identified by an integer 1 through N. M represents the number of the pairs of islands between which a bridge may be built. Each line of the next M lines contains three integers Si, Di and Ci (1 ≤ i ≤ M) which represent that it will cost Ci to build the bridge between islands Si and Di. You may assume 3 ≤ N ≤ 500, N − 1 ≤ M ≤ min(50000,N(N − 1)/2), 1 ≤ Si < Di ≤ N, and 1 ≤ Ci ≤ 10000. No two bridges connect the same pair of two islands, that is, if i ̸= j and Si = Sj, then Di ̸= Dj. If all the candidate bridges are built, all the islands are reachable from any other islands via one or more bridges.
Output
For each test case, output two integers, which mean the number of no alternative bridges and the sum of their construction cost, separated by a space.
Sample Input
4 4 1 2 3 1 3 3 2 3 3 2 4 3 4 4 1 2 3 1 3 5 2 3 3 2 4 3 4 4 1 2 3 1 3 1 2 3 3 2 4 3 3 3 1 2 1 2 3 1 1 3 1
Sample Output
1 3 3 9 2 4 0 0

分析:

题意
给定一个N个点,M条边的简单连通无向图。
对于一个无向图来说,它的最小生成树可能不是唯一的。
问在它的所有的最小生成树中共有的边是哪几条,输出边数和权值之和。
3<=N<=500, N-1<=M<=min{50000, N(N-1)/2}
思路
首先跑一遍Kruskal,得到最小生成树的权值。
之后尝试删去图中的边,如果某一条边被删去后,最小生成树的值发生了变化(一定变大),那么说明这条边是在所有的最小生成树中都不可或缺的,那么就把这条边加入到答案中。
注意到第一次Kruskal得到的边已经包含了所有的答案,因此只要枚举这里的N-1条边即可。
边排序的复杂度被均摊了,并查集的复杂度可以忽略,因此总的复杂度是O(NM)

注意:重载运算符比cmp快一点

还有就是用一个数组存下第一次MST用到的边(开始没有存起来,直接标记,超时。。。。。。。。。)

code:

#include <iostream>
#include <cstdio>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<math.h>
#include<memory>
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
#define max_v 50010
#define max_n 510
struct edge
{
int x,y,w;
bool operator<(const edge& b) const
{
return w<b.w;
}
} e[max_v];
int possEdge[max_n];//保存第一次MST用到的边 很重要
int pa[max_n];
int cnt,minv;
int n,m;
int c,wsum;
void init()
{
for(int i=;i<=n;i++)
pa[i]=i;
}
int find_set(int x)
{
if(x!=pa[x])
pa[x]=find_set(pa[x]);
return pa[x];
}
int union_set(int x,int y)
{
x=find_set(x);
y=find_set(y);
if(x==y)
return ;
pa[x]=y;
return ;
}
void firstkrus()
{
cnt=;
minv=;
init();
for(int i=; i<m; i++)
{
if(union_set(e[i].x,e[i].y))
{
minv+=e[i].w;
possEdge[cnt++]=i;
if(cnt==n-)
break;
}
}
}
bool krusWithout(int ce)
{
init();
int sum=,ct=;
for(int i=; i<m; i++)
{
if(i==ce)
continue;
if(union_set(e[i].x,e[i].y))
{
ct++;
sum+=e[i].w;
if(sum>minv)
return false;
if(ct==cnt)
return true;
}
}
return false;
}
void trycut()
{
c=;
wsum=;
for(int i=; i<cnt; i++)
{
if(!krusWithout(possEdge[i]))
{
c++;
wsum+=e[possEdge[i]].w;
}
}
printf("%d %d\n",c,wsum);
}
int main()
{
while(~scanf("%d %d",&n,&m))
{
for(int i=; i<m; i++)
scanf("%d %d %d",&e[i].x,&e[i].y,&e[i].w);
sort(e,e+m);
firstkrus();
trycut();
}
return ;
}

UVALive - 6837 Kruskal+一点性质(暴力枚举)的更多相关文章

  1. Gym 101194L / UVALive 7908 - World Cup - [三进制状压暴力枚举][2016 EC-Final Problem L]

    题目链接: http://codeforces.com/gym/101194/attachments https://icpcarchive.ecs.baylor.edu/index.php?opti ...

  2. P1217 [USACO1.5]回文质数 Prime Palindromes(技巧+暴力枚举+线性筛)

    技巧:就是偶数位的回文数字一定不是质数---------证明:奇数位之和sum1==偶数位之和sum2的数字可以被11整除.(11除外,这是一个坑点) 最高位,最低位必须是 1, 3, 7, 9 暴力 ...

  3. UVA.12716 GCD XOR (暴力枚举 数论GCD)

    UVA.12716 GCD XOR (暴力枚举 数论GCD) 题意分析 题意比较简单,求[1,n]范围内的整数队a,b(a<=b)的个数,使得 gcd(a,b) = a XOR b. 前置技能 ...

  4. Codeforces Round #266 (Div. 2)B(暴力枚举)

    很简单的暴力枚举,却卡了我那么长时间,可见我的基本功不够扎实. 两个数相乘等于一个数6*n,那么我枚举其中一个乘数就行了,而且枚举到sqrt(6*n)就行了,这个是暴力法解题中很常用的性质. 这道题找 ...

  5. CodeForces 742B Arpa’s obvious problem and Mehrdad’s terrible solution (暴力枚举)

    题意:求定 n 个数,求有多少对数满足,ai^bi = x. 析:暴力枚举就行,n的复杂度. 代码如下: #pragma comment(linker, "/STACK:1024000000 ...

  6. 2014牡丹江网络赛ZOJPretty Poem(暴力枚举)

    /* 将给定的一个字符串分解成ABABA 或者 ABABCAB的形式! 思路:暴力枚举A, B, C串! */ 1 #include<iostream> #include<cstri ...

  7. HNU 12886 Cracking the Safe(暴力枚举)

    题目链接:http://acm.hnu.cn/online/?action=problem&type=show&id=12886&courseid=274 解题报告:输入4个数 ...

  8. 51nod 1116 K进制下的大数 (暴力枚举)

    题目链接 题意:中文题. 题解:暴力枚举. #include <iostream> #include <cstring> using namespace std; ; ; ch ...

  9. Codeforces Round #349 (Div. 1) B. World Tour 最短路+暴力枚举

    题目链接: http://www.codeforces.com/contest/666/problem/B 题意: 给你n个城市,m条单向边,求通过最短路径访问四个不同的点能获得的最大距离,答案输出一 ...

随机推荐

  1. OpenStack IceHouse 部署 - 2 - 网络与软件环境初始化

    OpenStack应用:节点软硬件环境配置    节点硬件与IP分配 实验室网关 10.14.39.1 各个节点 节点名称 硬件(Linux硬盘分区,RAM,CPU) ip地址(接口) 作用与运行的服 ...

  2. 思维导图_Python_内置函数

  3. 007Spring Security

    01.基于Spring AOP 和 Servlet规范中Filter实现  的安全框架 <dependency> <groupId>org.springframework.se ...

  4. 4类Storage方案(AS开发实战第四章学习笔记)

    4.1 共享参数SharedPreferences SharedPreferences按照key-value对的方式把数据保存在配置文件中,该配置文件符合XML规范,文件路径是/data/data/应 ...

  5. jquery 之节点操作

    一.添加节点 [添加内部子节点方法]:内部节点就是儿子节点 append()    在被选元素内部的结尾插入内容 appendTo()  将指定内容插入到被选标签内部的结尾 prepend()   在 ...

  6. JavaScript停止事件冒泡和取消事件默认行为

    功能:停止事件冒泡 function stopBubble(e) { // 如果提供了事件对象,则这是一个非IE浏览器 if ( e && e.stopPropagation ) { ...

  7. mysql 的增删改查

    数据库的基本流程就是先看你的数据库中的库都是哪些:show databases; 然后再进入相应的库进行操作  :  use+进入的库/表 切换路径 查看这个库内的所有的表: show tabales ...

  8. 编码学习---代码OJ网站

    代码OJ网站: https://leetcode-cn.com/accounts/login/

  9. Office 365实现单点登录系列(1)—域环境搭建

    Hello 小伙伴们, 2018新年快乐,作为2018年首篇文章,怎么能不给大家带来点干货呢?这篇文章其实我9月底的时候已经在MSDN上发布过了,为表诚意,我更新了这篇文章,并把它组成了一个系列,2. ...

  10. TFS使用笔记

    TFS是用来存储文件的服务器,放置不同版本的文件.因此文件的数量和内容因版本不同而不同. 在Fig-00中,服务器和本地的对应文件夹Common下的因版本不一致,所以文件数量是不一致的. Fig-00 ...