POJ1679--The Unique MST（次小生成树）

最新推荐文章于 2019-05-04 17:33:50 发布

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本文介绍了一种算法，用于判断给定的无向图是否存在唯一的最小生成树（MST）。通过构造次小生成树并与最小生成树进行比较来实现这一目标。文章详细解释了算法流程，包括使用广度优先搜索确定MST上任意两点间路径的最大权重。

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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

Sample Output

3
Not Unique!

题意：询问最小生成树是否唯一。只需求出次小生成树后与之比较即可

求次小生成树：先求最小生成树后枚举每条没有用的边，添加此边必然构成一个环。删除环中权最大的边即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 180
#define maxm 10080
#define inf 0x3f3f3f3f
int first[maxn];
int vv[maxm],nxt[maxm],ww[maxm];
int mcost[maxn][maxn];
int e,n,m;
int father[maxn];
bool vis[maxm];

void AddEdge(int u,int v,int w)
{
	vv[e] = v;	nxt[e] = first[u];	ww[e] = w;	first[u] = e++;
	vv[e] = u;	nxt[e] = first[v];	ww[e] = w;	first[v] = e++;
}

void init()
{
	e = 0;
	memset(first,-1,sizeof(first));
	memset(vis,0,sizeof(vis));//边
	memset(mcost,0,sizeof(mcost));
	for(int i = 1;i <= n;i++)	father[i] = i;
}

int find(int x)
{
	if(x == father[x])	return x;
	return father[x] = find(father[x]);
}

void Union(int u,int v)
{
	father[u] = v;
}

struct Edge
{
	int u,v,w;
	bool operator < (const Edge & a) const
	{
		return w < a.w;
	}
}edge[maxm];

int Build_Mst()
{
	int mst = 0;
	for(int i = 1;i <= m;i++)
	{	
		int u = edge[i].u,v = edge[i].v,w = edge[i].w;	
		int fa = find(u),fb = find(v);	
		if(fa == fb)	continue;
		Union(fa,fb);	vis[i] = 1;
		mst += w;
		AddEdge(u,v,w);
	}
	return mst;
}

int Build_Sst()
{
	int add = inf;
	for(int i = 1;i <= m;i++)
	{
		if(!vis[i])
		{
			int u = edge[i].u,v = edge[i].v,w = edge[i].w;
			add = min(add,w - mcost[u][v]);
		}
	}
	return add;
}

void bfs(int s,int u,int fa,int mc)
{
	for(int i = first[u];i != -1;i = nxt[i])
	{
		int v = vv[i],w = ww[i];
		if(v == fa)	continue;
		if(w > mc)	mcost[s][v] = w;
		else mcost[s][v] = mc;
		bfs(s,v,u,mcost[s][v]);
	}
}

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		init();
		for(int i = 1;i <= m;i++)
			scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
		sort(edge+1,edge+1+m);
		int mst = Build_Mst();
		for(int i = 1;i <= n;i++)	bfs(i,i,-1,0);//广搜确定MST上任意两点路径上的最长路
		int sst = Build_Sst();
		if(sst == 0)	puts("Not Unique!");
		else printf("%d\n",mst);
	}
	return 0;
}