The Unique MST

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!

题意描述：
给出n个点及它们之间的路径判断它们的最小生成树是否唯一，若唯一输出结果，不唯一输出“Not Unique!”

解题思路：

先利用Kruskal算法求出最小生成树结果sum，将形成最小生成树的路径进行标记，再利用枚举法，每次去掉一个最小生成树的边，用其他边代替，若能找到最小生成树的结果还为sum即说明最小生成树不唯一；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int f[1010],n,m;
struct A
{
	int a;
	int b;
	int c;
	int h;
}q[100010];
int cmp(struct A x,struct A y)
{
	return x.c<y.c;
}
int get(int i)
{
	if(f[i]==i)
		return i;
	else
	{
		f[i]=get(f[i]);
		return f[i];
	}
}
int merge(int u,int v)
{
	int t1;
	int t2;
	t1=get(u);
	t2=get(v);
	if(t1!=t2)
	{
		f[t2]=t1;
		return 1;
	}
	return 0;
}
int kush(int x)
{
	int i,sum,cut;
	for(i=1;i<=n;i++)
		f[i]=i;
	sum=0;
	cut=0;
	for(i=1;i<=m;i++) 
	{
		if(i==x)//跳过这条已用过的边 
			continue;
		if(merge(q[i].a,q[i].b))
		{
			sum=sum+q[i].c;
			cut++;
		}
		if(cut==n-1)
			return sum;
	}
	return -1;
}
int main()
{
	int t,i,j,sum,cut,ans,flag;
	while(scanf("%d",&t)!=EOF)
	{
		while(t--)
		{
			scanf("%d%d",&n,&m);
			for(i=1;i<=m;i++)
			{
				scanf("%d%d%d",&q[i].a,&q[i].b,&q[i].c);
				q[i].h=0;
			}
			sort(q+1,q+m+1,cmp);
			for(i=1;i<=n;i++)
				f[i]=i;
			sum=0;
			cut=0;
			//Kruskal
			for(i=1;i<=m;i++)
			{
				if(merge(q[i].a,q[i].b))
				{
					sum=sum+q[i].c;
					q[i].h=1;
					cut++;
				}
				if(cut==n-1)
					break;
			}
			flag=0;
			for(i=1;i<=m;i++)
			{
				if(q[i].h==1)//枚举每次减去一条边 
					ans=kush(i);
				if(sum==ans)//若相等说明最小生成树已不是唯一，结束循环 
				{
					flag=1;
					break;
				}
			}
			if(flag==1)
				printf("Not Unique!\n");
			else
				printf("%d\n",sum);
		}
	}
	return 0;
}