Given a connected undirected graph, tell if its minimum spanning tree is unique.

Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties: 
1. V' = V. 
2. T is connected and acyclic.

Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'.

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.

Output

For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.

Sample Input

2
3 3
1 2 1
2 3 2
3 1 3
4 4
1 2 2
2 3 2
3 4 2
4 1 2

Sample Output

3
Not Unique! 求次小生成树 看与最小生成树是否相同
prime求次小生成树
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = , INF = 0x7fffffff;
typedef long long LL;
int graph[][], d[maxn], vis[maxn], maxd[][], pre[maxn];
int n, m; int prime(int s)
{
int temp, sum = ;
mem(vis, );
for(int i=; i<=n; i++) d[i] = graph[s][i], pre[i] = s;
vis[s] = ;
d[s] = ;
for(int i=; i<n; i++)
{
int mincost = INF;
for(int j=; j<=n; j++)
{
if(!vis[j] && mincost > d[j])
mincost = d[j], temp = j;
}
for(int j=; j<=n; j++)
if(vis[j]) maxd[temp][j] = maxd[j][temp] = max(mincost, maxd[pre[temp]][j]);
vis[temp] = ;
sum += mincost;
for(int j=; j<=n; j++)
{
if(!vis[j] && d[j] > graph[temp][j])
d[j] = graph[temp][j], pre[j] = temp;
}
}
// for(int i=1; i<=n; i++)
// sum += d[i];
return sum;
} int main()
{
int T;
cin>> T;
while(T--)
{
cin>> n >> m;
for(int i=; i<=n; i++)
for(int j=; j<=n; j++)
if(i == j) graph[i][j] = ;
else graph[i][j] = graph[j][i] = INF;
for(int i=; i<m; i++)
{
int u, v, w;
cin>> u >> v >> w;
graph[u][v] = graph[v][u] = w;
}
int sum = prime();
int lsum = INF;
for(int i=; i<=n; i++)
for(int j=i+; j<=n; j++)
{
if(i != pre[j] && j != pre[i] && graph[i][j] != INF)
if(sum - maxd[i][j] + graph[i][j] < lsum)
lsum = sum - maxd[i][j] + graph[i][j];
} if(lsum == sum)
cout<< "Not Unique!" <<endl;
else
cout<< sum <<endl; } return ;
}
												

The Unique MST POJ - 1679 (次小生成树)的更多相关文章

  1. The Unique MST POJ - 1679 次小生成树prim

    求次小生成树思路: 先把最小生成树求出来  用一个Max[i][j] 数组把  i点到j 点的道路中 权值最大的那个记录下来 used数组记录该条边有没有被最小生成树使用过   把没有使用过的一条边加 ...

  2. Day5 - G - The Unique MST POJ - 1679

    Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spann ...

  3. POJ 1679 The Unique MST 【最小生成树/次小生成树模板】

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 22668   Accepted: 8038 D ...

  4. poj1679 The Unique MST(判定次小生成树)

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23180   Accepted: 8235 D ...

  5. POJ-1679.The Unique MST.(Prim求次小生成树)

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 39561   Accepted: 14444 ...

  6. poj 1679 次小生成树

    次小生成树的求法: 1.Prime法 定义一个二维数组F[i][j]表示点i到点j在最小生成树中的路径上的最大权值.有个知识就是将一条不在最小生成树中的边Edge加入最小生成树时,树中要去掉的边就是E ...

  7. K - The Unique MST - poj 1679

    题目的意思已经说明了一切,次小生成树... ****************************************************************************** ...

  8. (最小生成树 次小生成树)The Unique MST -- POJ -- 1679

    链接: http://poj.org/problem?id=1679 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82831#probl ...

  9. The Unique MST POJ - 1679 最小生成树判重

    题意:求一个无向图的最小生成树,如果有多个最优解,输出"Not Unique!" 题解: 考虑kruskal碰到权值相同的边: 假设点3通过边(1,3)连入当前所维护的并查集s. ...

随机推荐

  1. Python内存优化:Profile,slots,compact dict

    实际项目中,pythoner更加关注的是Python的性能问题,之前也写过一篇文章<Python性能优化>介绍Python性能优化的一些方法.而本文,关注的是Python的内存优化,一般说 ...

  2. [Oracle][DataGuard]Standby数据库文件有损坏时的处理方法

    需要参考: [Oracle]Recovering the primary database's datafile using the physical standby, and vice versa ...

  3. aurora 64B/66B ip核设置与例程代码详解

    见网页https://blog.csdn.net/u014586651/article/details/84349328 https://blog.csdn.net/u012135070/articl ...

  4. 我的devops实践经验分享一二

    前言 随着系统越来越大,开发人员.站点.服务器越来越多,微服务化推进,......等等原因,实现自动化的devops越来越有必要. 当然,真实的原因是,在团队组建之初就预见到了这些问题,所以从一开始就 ...

  5. c#代码分析

    代码分析是在一个IT行业计算机程序员必须要具有的基本专业技能,在了解一定的专业基础之上去看懂别人编写的代码,分析别人代码实现的功能,以及基本的维护和扩展测试.不同的人有不同的代码风格,要使自己的能要别 ...

  6. sring引入mybatis

    1.首先框架结构是这样的(jar包还是要导的) 2.web.xml和springMVC-servlet.xml未作任何新的配置,这里简单贴一下代码: <?xml version="1. ...

  7. 安装python包时报错

    pip install numpy  时  报错: Traceback (most recent call last):  File "d:\学习\python\python-3.6.5\l ...

  8. Python学习笔记——Python Number(数字)

    Python Number 类型转换 int(x, y) #将x转换为一个整数,y为进制数.如 int('11',2)将二进制数的11转成十进制数的整数,结果为3 long(x, y) #将x转换为一 ...

  9. solr6.2单机版安装

    1安装solr服务,先安装jdk和tomcat 2去官网(http://archive.apache.org/dist/lucene/solr/)下载solr压缩包,最新版本是6.4.1,下载解压后, ...

  10. [2017BUAA软工]第0次博客作业

    第一部分:结缘计算机 1.你为什么选择计算机专业?你认为你的条件如何?和这些博主比呢? 当初选择计算机专业作为自己报考大学的第一志愿,主要是看重了市场对于计算机行业人士的巨大需求,同时也感慨于计算机行 ...