The Unique MST

Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 36692   Accepted: 13368

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! 题解:
次小生成树,维护一个两点间的最小距离,最后再向上加
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
#define line cout<<"------------------"<<endl;
const int MAXN=1e4+10;
const int INF=0x3f3f3f3f;
int n,m;
struct node{
int x,y;
int v;
bool vis;
}Edge[MAXN];
bool cmp(node a,node b)
{
return a.v<b.v;
}
int pre[MAXN];
int Find(int a)
{
if(pre[a]==a)
return a;
return Find(pre[a]);
}
vector<int >G[110]; int maxd[110][110];//并查集划到一个树上后,树上任意两点之间的距离 void init()
{
for (int i = 1; i <=n; ++i) {
G[i].clear();
pre[i] = i;
G[i].push_back(i);
} }
int main()
{
int _;
scanf("%d",&_);
while(_--)
{
scanf("%d%d",&n,&m);
init();
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&Edge[i].x,&Edge[i].y,&Edge[i].v);
Edge[i].vis=false;
}
sort(Edge+1,Edge+1+m,cmp);
int sum=0;
for (int i = 1; i <=m ; ++i) {
int x=Find(Edge[i].x);
int y=Find(Edge[i].y);
if(x!=y)
{
pre[x]=y;
sum+=Edge[i].v;
int len1=G[x].size();
int len2=G[y].size();
for (int j = 0; j <len1 ; ++j) {
for (int k = 0; k <len2 ; ++k) {
maxd[G[x][j]][G[y][k]]=maxd[G[y][k]][G[x][j]]=Edge[i].v;//构建两点间最小距离
}
}
int tem[110];
for (int j = 0; j <len2 ; ++j) {
tem[j]=G[y][j];
}
for (int j = 0; j <len1 ; ++j) {
G[y].push_back(G[x][j]);
}
for (int j = 0; j <len2 ; ++j) {
G[x].push_back(tem[j]);
}
Edge[i].vis=true;
}
}
int cis=INF;
for (int i = 1; i <=m ; ++i) {//从不是最小生成树上的边,遍历向上加。找到次小生成树
if(!Edge[i].vis)
cis=min(cis,sum+Edge[i].v-maxd[Edge[i].x][Edge[i].y]);
}
if(cis>sum)
printf("%d\n",sum);
else
printf("Not Unique!\n");
} return 0;
}
//poj1679

  

POJ1679(次小生成树)的更多相关文章

  1. POJ1679(次小生成树)

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24201   Accepted: 8596 D ...

  2. poj1679次小生成树入门题

    次小生成树求法:例如求最小生成树用到了 1.2.4这三条边,总共5条边,那循环3次的时候,每次分别不用1.2.4求得最小生成树的MST,最小的MST即为次小生成树 如下代码maxx即求最小生成树时求得 ...

  3. poj1679 次小生成树

    prim方法:先求过一遍prim,同时标记使用过得边.然后同时记录任意2点间的最大值. 每次加入一条新的边,会产生环,删去环中的最大值即可. #include<stdio.h> #incl ...

  4. POJ1679 The Unique MST[次小生成树]

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

  5. 次小生成树(POJ1679/CDOJ1959)

    POJ1679 首先求出最小生成树,记录权值之和为MinST.然后枚举添加边(u,v),加上后必形成一个环,找到环上非(u,v)边的权值最大的边,把它删除,计算当前生成树的权值之和,取所有枚举加边后生 ...

  6. POJ1679 The Unique MST【次小生成树】

    题意: 判断最小生成树是否唯一. 思路: 首先求出最小生成树,记录现在这个最小生成树上所有的边,然后通过取消其中一条边,找到这两点上其他的边形成一棵新的生成树,求其权值,通过枚举所有可能,通过这些权值 ...

  7. POJ1679 The Unique MST 【次小生成树】

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

  8. 次小生成树(poj1679)

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20737   Accepted: 7281 D ...

  9. 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 ...

随机推荐

  1. 【Leetcode】【Easy】Binary Tree Level Order Traversal II

    Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...

  2. Ant Design项目记录和CSS3的总结和Es6的基本总结

    这里主要是介绍自己运用ANT框架的一些小总结,以前写到word里,现在要慢慢传上来, 辅助生殖项目总结:从每个组件的运用的方法和问题来总结项目. 1.项目介绍 辅助生殖项目主要运用的是Ant.desi ...

  3. 如何将一个PDF文件里的图片批量导出

    假设我有下面这个PDF文件,里面有很多图片,我想把这些图片批量导出,而不是在Adobe Acrobat Reader里一张张手动拷贝: 本文介绍一种快捷做法. 用PDF-XChange Editor打 ...

  4. Android(java)学习笔记57:PC and Phone 通信程序

    1. 首先我写的程序代码如下: package com.himi.udpsend; import java.net.DatagramPacket; import java.net.DatagramSo ...

  5. 编程思想的理解(POP,OOP,SOA,AOP) x

    http://blog.chinaunix.net/uid-29417436-id-4060980.html 1)POP--面向过程编程(Process-oriented programming ): ...

  6. CSU计算机研究生推免

    考研复习 一开始,我是没有想到能够拿到研究生推免资格的,从今年3月份到整个暑假过完,一共6个月的时间,我一直在准备考研. 具体来说,我是在准备考研数学,整整6个月时间的数学复习,给我一种感觉,把大一大 ...

  7. 2018_MCM_ICM_C

  8. vue案列

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

  9. HDU 1099 Lottery (求数学期望)

    传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1099 Lottery Time Limit: 2000/1000 MS (Java/Others)   ...

  10. hbase 数据拷贝

    由于运营数据太大,另外避免影响正常访问,所以需要临时拷贝部分数据到临时表中. bin/hbase org.apache.hadoop.hbase.mapreduce.CopyTable [--star ...