The Unique MST
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 35999   Accepted: 13145

Description

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!

C/C++:

 #include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
const int my_max_edge = , my_max_node = ; int t, n, m, my_book_edge[my_max_edge], my_pre[my_max_node], my_first; struct edge
{
int a, b, val;
}P[my_max_edge]; bool cmp(edge a, edge b)
{
return a.val < b.val;
} int my_find(int x)
{
int n = x;
while (n != my_pre[n])
n = my_pre[n];
int i = x, j;
while (n != my_pre[i])
{
j = my_pre[i];
my_pre[i] = n;
i = j;
}
return n;
} int my_kruskal(int my_flag)
{
int my_ans = ;
for (int i = ; i <= n; ++ i)
my_pre[i] = i; for (int i = ; i < m; ++ i)
{
int n1 = my_find(P[i].a), n2 = my_find(P[i].b);
if (n1 == n2 || my_flag == i) continue;
my_pre[n1] = n2;
if (my_first)my_book_edge[i] = ;
my_ans += P[i].val;
} int temp = my_find();
for (int i = ; i <= n; ++ i)
if (temp != my_find(i))
return -;
return my_ans;
} int main()
{
scanf("%d", &t);
while (t --)
{
scanf("%d%d", &n, &m);
for (int i = ; i < m; ++ i)
scanf("%d%d%d", &P[i].a, &P[i].b, &P[i].val);
sort(P, P + m, cmp);
memset(my_book_edge, , sizeof(my_book_edge)); my_first = ;
int mst = my_kruskal(-), flag = ;
if (mst == -)
{
printf("0\n");
continue;
}
my_first = ;
for (int i = ; i < m; ++ i)
{
if (my_book_edge[i])
{
if (mst == my_kruskal(i))
{
printf("Not Unique!\n");
flag = ;
break;
}
}
}
if (flag) printf("%d\n", mst);
}
return ;
}

poj 1679 The Unique MST (次小生成树(sec_mst)【kruskal】)的更多相关文章

  1. POJ 1679 The Unique MST (次小生成树)

    题目链接:http://poj.org/problem?id=1679 有t组数据,给你n个点,m条边,求是否存在相同权值的最小生成树(次小生成树的权值大小等于最小生成树). 先求出最小生成树的大小, ...

  2. POJ 1679 The Unique MST (次小生成树 判断最小生成树是否唯一)

    题目链接 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. De ...

  3. POJ 1679 The Unique MST (次小生成树kruskal算法)

    The Unique MST 时间限制: 10 Sec  内存限制: 128 MB提交: 25  解决: 10[提交][状态][讨论版] 题目描述 Given a connected undirect ...

  4. poj 1679 The Unique MST 【次小生成树】【模板】

    题目:poj 1679 The Unique MST 题意:给你一颗树,让你求最小生成树和次小生成树值是否相等. 分析:这个题目关键在于求解次小生成树. 方法是,依次枚举不在最小生成树上的边,然后加入 ...

  5. 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 ...

  6. POJ1679 The Unique MST —— 次小生成树

    题目链接:http://poj.org/problem?id=1679 The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total S ...

  7. poj 1679 The Unique MST

    题目连接 http://poj.org/problem?id=1679 The Unique MST Description Given a connected undirected graph, t ...

  8. poj 1679 The Unique MST(唯一的最小生成树)

    http://poj.org/problem?id=1679 The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submis ...

  9. poj 1679 The Unique MST (判定最小生成树是否唯一)

    题目链接:http://poj.org/problem?id=1679 The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total S ...

随机推荐

  1. iguana::json/xml 序列化框架

    环境:win10  vs2017  c++17  boost 1.下载源码:https://github.com/qicosmos/iguana 2.创建工程,包含源码目录.boost库目录:boos ...

  2. python中的__call__函数

    简单实例: class TmpTest: def __init__(self, x, y): self.x = x self.y = y def __call__(self, x, y): self. ...

  3. ThreadPoolExecutor使用方法

    先看构造方法 ,ThreadPoolExecutor共4个构造方法: 直接看参数最多的7个参数分别代表: public ThreadPoolExecutor(int corePoolSize, int ...

  4. VBA 在第二个sheet中查找第一个sheet中不存在的值

    VBA 在第二个sheet中查找第一个sheet中不存在的值  Sub Macro2() ' ' Macro2 Macro ' 宏由 Lizm 录制,时间: 2019/04/10 '   ' Dim ...

  5. java类在何时被加载

    我们接着上一章的代码继续来了解一下java类是在什么时候加载的.在开始验证之前,我们现在IDEA做如下配置. -XX:+TraceClassLoading 监控类的加载 我们新建了一个TestCont ...

  6. jquery链式原理.html

    <!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...

  7. GO基础之List

    一.List定义 概述1.list是一种非连续存储的容器,由多个节点组成,节点通过一些变量记录彼此之间的关系.list有多种实现方法,如单向链表.双向链表等.2.Go语言中list的实现原理是双向链表 ...

  8. SpringBoot之响应式编程

    一 Spring WebFlux Framework说明 Spring WebFlux 是 Spring Framework 5.0 中引入的新 reactive web framework.与 Sp ...

  9. Unity中的资源管理

    一.AssetBundle 相关 Q1:Unity中的SerializedFile是怎么产生的?请问用Unload(false)可以清除吗?因为读取了Bundle里面的内容后已经赋值给其他物体了.而且 ...

  10. 一道国外前端面试题引发的Coding...

    刚刚看到CSDN微信公众号一篇文章,关于国外程序员面试前端遇到的一道测试题,有点意思,遂写了下代码,并记录一下~ 题目是这样的: ['Tokyo', 'London', 'Rome', 'Donlon ...