http://poj.org/problem?id=1679

Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 30120   Accepted: 10778

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!

Source

 
 
蒟蒻就是弱、、、
 #include <algorithm>
#include <cstdio> using namespace std; const int N();
int num,n,m;
int fa[N],cnt;
int Fir,MST;
int u,v,w,used[N];
struct Edge
{
int u,v,w;
} edge[N<<]; bool cmp(Edge a,Edge b)
{
return a.w<b.w;
} int find(int x)
{
return fa[x]==x?x:fa[x]=find(fa[x]);
} int Kruskal()
{
int ans=; cnt=;
sort(edge+,edge+m+,cmp);
for(int i=; i<=n; i++) fa[i]=i;
for(int i=; i<=m; i++)
{
int fx=find(edge[i].u),fy=find(edge[i].v);
if(fx!=fy)
{
fa[fx]=fy;
used[++cnt]=i;
ans+=edge[i].w;
}
if(cnt==n-) return ans;
}
return ans;
} int SecKru(int cant)
{
int ans=; cnt=;
for(int i=;i<=n;i++) fa[i]=i;
for(int i=;i<=m;i++)
{
if(cant==i) continue;
int fx=find(edge[i].u),fy=find(edge[i].v);
if(fx!=fy)
{
cnt++;
fa[fx]=fy;
ans+=edge[i].w;
}
if(cnt==n-) return ans;
}
return 0x7fffffff;
} int main()
{
scanf("%d",&num);
for(;num--;)
{
scanf("%d%d",&n,&m);
for(int i=;i<=m;i++)
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
Fir=Kruskal(); MST=0x7fffffff;
for(int i=;i<n;i++)
{
MST=min(SecKru(used[i]),MST);
}
if(Fir==MST)
printf("Not Unique!\n");
else printf("%d\n",Fir);
}
return ;
}

POJ——T1679 The Unique MST的更多相关文章

  1. 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 ...

  2. 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 ...

  3. POJ 1679 The Unique MST (最小生成树)

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

  4. 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 ...

  5. 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 ...

  6. POJ 1679 The Unique MST (最小生成树)

    The Unique MST 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/J Description Given a conn ...

  7. poj 1679 The Unique MST【次小生成树】

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24034   Accepted: 8535 D ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. 多个API接口

    青云客智能聊天机器人API 为保证接口稳定,API接口正式启用 api.qingyunke.com,www.qingyunke.com 如何更换API接口地址? 答:如果您的API接口是用在微信公众号 ...

  2. CF899A Splitting in Teams

    CF899A Splitting in Teams 题意翻译 n个数,只有1,2,把它们任意分组,和为3的组最多多少 题目描述 There were nn groups of students whi ...

  3. 【配置属性】—Entity Framework实例详解

    Entity Framework Code First的默认行为是使用一系列约定将POCO类映射到表.然而,有时候,不能也不想遵循这些约定,那就需要重写它们.重写默认约定有两种方式:Data Anno ...

  4. HDU 4355

    只能说感觉是三分吧,因为两端值肯定是最大的,而中间肯定存在一点使之最小,呃,,,,猜 的... #include <iostream> #include <cstdio> #i ...

  5. 自己定义View实现水平滚动控件

    前几天项目中须要使用到一个水平可滚动的选择条,类似下图效果(图片是从简书上一位作者那儿找来的,本篇也是在这位作者的文章的基础上改动的,站在大神的肩膀上,哈哈,因为原文没有提供demo,并且实现的效果跟 ...

  6. php+mysql 最简单的留言板

    学完了记得动手操作. 測试地址(未过滤) <html> <body> <head><meta http-equiv="Content-Type&qu ...

  7. 项目记录22-- tolua基于lua框架事件派发

     每天晚上抽点时间写一点点就一点点,曾经不写博客可是如今.不为别的仅仅是为了告诉别人我还存在.         这几天在地铁上发现好多人都还在玩消除游戏,今天起码看到5个人,可是玩的版本号都不一样.看 ...

  8. 浅析PHP中cookie与session技术

    1.cookie是什么? cookie指某些站点为了辨别用户身份.进行session跟踪而储存在用户本地终端上的数据(通常经过加密). 通俗来理解就是,你去一个专卖店或者超市买东西,然后店里给你办一张 ...

  9. 基于macOS+VMware的GNS3内VM上公网

    笔者经常需要做网络实验,GNS3就是笔者最喜欢用的模拟器,为了便于实验,需要能从macos上直接ssh登陆模拟出来的vm,并且vm需要上公网.经过研究,已解决此问题,并以此分享出来 tag: maco ...

  10. Linux,Docker,Jenkins No such file or directory

    你们先休息下,我先哭哭! 今天在做交接项目的bug修改的时候,在创建文件的时候报错 No such file or directory 然后跟着路径去linux中查看了该路径,但确实存在,并且权限都是 ...