The Unique MST
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 28207   Accepted: 10073

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

 
求一个图是否有不同的最小生成树
大家都太暴力,枚举去那条边,暴力一遍判断重复
但其实可以更快
可以参考这篇论文(注意,文章中代码我认为有错,请自行思考)
http://www.docin.com/p-806495282.html
 #include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<climits>
#include<algorithm>
#include<queue>
#define LL long long
using namespace std;
typedef struct{
int to,frm,dis;
}edge;
edge gra[];
int num=,fa[];
int n,m;
int cmp(const edge &a,const edge &b){
return a.dis<b.dis;
}
int fnd(int x){
return x==fa[x]?x:fnd(fa[x]);
}
int uni(int x,int y){
int fx=fnd(x);
int fy=fnd(y);
fa[fy]=fx;
return ;
}
inline int read(){
int sum=;char ch=getchar();
while(ch>''||ch<'')ch=getchar();
while(ch<=''&&ch>=''){
sum=sum*+ch-'';
ch=getchar();
}
return sum;
}
int kru(){
int ans=;
sort(gra+,gra+m+,cmp);
for(int i=;i<=n;i++)fa[i]=i;
for(int i=;i<=m;i++){
int x=gra[i].frm;
int y=gra[i].to;
int fx=fnd(x);
int fy=fnd(y);
if(fx!=fy){
int j=i+;
while(j<=m&&gra[j].dis==gra[i].dis){
int y1=gra[j].frm;
int x1=gra[j].to;
int fy1=fnd(y1);
int fx1=fnd(x1);
if((fx1==fx&&fy1==fy)||(fx1==fy&&fy1==fx))return -;
j++;
}
ans+=gra[i].dis;
uni(fx,fy);
}
}
return ans;
}
int main(){
int t;
t=read();
while(t--){
memset(gra,,sizeof(gra));
n=read(),m=read();
num=;
for(int i=;i<=m;i++){
gra[i].frm=read();
gra[i].to=read();
gra[i].dis=read();
} int ans=kru();
if(ans==-)printf("Not Unique!\n");
else printf("%d\n",ans);
}
return ;
}

[poj1679]The Unique MST(最小生成树)的更多相关文章

  1. POJ1679 The Unique MST(Kruskal)(最小生成树的唯一性)

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 27141   Accepted: 9712 D ...

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

    http://poj.org/problem?id=1679 Description Given a connected undirected graph, tell if its minimum s ...

  3. POJ1679 The Unique MST[次小生成树]

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28673   Accepted: 10239 ...

  4. POJ1679 The Unique MST 【次小生成树】

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20421   Accepted: 7183 D ...

  5. POJ1679 The Unique MST 2017-04-15 23:34 29人阅读 评论(0) 收藏

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 29902   Accepted: 10697 ...

  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,次小生成树模板题

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K       Description Given a connected undirec ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. leetcode 419

    题目说明: Given an 2D board, count how many different battleships are in it. The battleships are represe ...

  2. jsp页面传参大汇总-转帖收藏

    http://blog.csdn.net/ssy_shandong/article/details/9328985/

  3. Hosts文件

    Hosts是一个没有扩展名的系统文件,可以用记事本等工具打开, 其作用:就是将一些常用的网址域名与其对应的IP地址建立一个关联"数据库",当用户在浏览器中输入一个需要登录的网址时, ...

  4. linux-------------计划任务crond:如何创建linux里面的计划任务

    1.centos下安装crond [root@localhost /]# yum -y install vixie-cron [root@localhost /]# yum -y install cr ...

  5. SetSysColors 修改系统颜色

    首先我们来看一下SetSysColors函数的原型: BOOL WINAPI SetSysColors( __in int cElements, //要改变的对象的数量 __in const INT* ...

  6. 开始学习bizTalk server了

    开始学习bizTalk Server 2013 R2了,有兴趣的朋友可以关注我,一同学习

  7. http 中定义的八种请求的介绍

    在http1.1协议中,共定义了8种可以向服务器发起的请求(这些请求也叫做方法或动作),本文对这八种请求做出简要的介绍: 1.PUT:put的本义是推送 这个请求的含义就是推送某个资源到服务器,相当于 ...

  8. 微信接口php

    官方提供的SDK只有一个文本消息功能,我们将所有消息的消息类型及事件响应都整理了进来,并且加入日志记录,代码如下: 更新日志: 2013-01-01 版本1.0,包含Token验证及基本消息接口的收发 ...

  9. js'中的apply和call和bind的用法

    apply:方法能劫持另外一个对象的方法,继承另外一个对象的属性. Function.apply(obj,args)方法能接收两个参数obj:这个对象将代替Function类里this对象args:这 ...

  10. Makefile编译库

    funs.h: #ifndef __FUNS_H__ #define __FUNS_H__ void fun1(); #endif funs.c #include "funs.h" ...