The Unique MST_生成树2018_3_5

最新推荐文章于 2021-07-27 19:04:04 发布

ujn20161222

最新推荐文章于 2021-07-27 19:04:04 发布

阅读量237

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/ujn20161222/article/details/79450612

poj 同时被 2 个专栏收录

130 篇文章

订阅专栏

生成树

5 篇文章

订阅专栏

本文介绍了一种算法，用于判断给定的带权无向图的最小生成树是否唯一。通过输入节点数、边数及各边权重，该算法能够确定最小生成树的总成本并检查其唯一性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

3
Not Unique!

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=105;
const int inf=1<<30;
int g[N][N],dis[N];
bool vis[N];
int main(){
	int t;
	scanf("%d",&t);
	int n,m,ans;
	while(t--){
		fill(&g[0][0],&g[0][0]+N*N,inf);
		scanf("%d%d",&n,&m);
		while(m--){
			int u,v,d;
			scanf("%d%d%d",&u,&v,&d);
			g[u][v]=g[v][u]=d;
		}
		memset(vis,0,sizeof(vis));
		for(int i=1;i<=n;i++)
		dis[i]=g[1][i];
		vis[1]=1;
		int u;
		ans=0;
		int i;
		for(i=1;i<n;i++){
			int mm=inf;
			for(int j=1;j<=n;j++)
			if(!vis[j]&&dis[j]<mm){
				mm=dis[j];u=j;
			}
			if(mm==inf)break;
			int num=0;
			for(int j=1;j<=n;j++)
			if(g[j][u]==mm&&vis[j])
			num++;
			if(num>1)break;
			vis[u]=1;
			ans+=mm;
			for(int j=1;j<=n;j++)
			if(!vis[j]&&g[j][u]<dis[j])
			dis[j]=g[j][u];
		}
		if(i==n)printf("%d\n",ans);
		else puts("Not Unique!");
	}
}