The Unique MST

Title 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!

The main idea of the topic

Ask if the minimum spanning tree of this graph is unique .
To judge whether it is unique, we only need to compare its minimum spanning tree and non strict sub small spanning tree .
This problem is made for and non strict sub small spanning trees .

When we link , Make a note of , The weight of each element of two sets to each other . Because if you want to replace one side , We must know the weight of the replaced edge .
Well, the core part explains , Then continue to

Accepted Code

//#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<string.h>
#include <iomanip>
#include<stdio.h>
#include<vector>
#include<string>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll ll_inf = 9223372036854775807;
const int int_inf = 2147483647;
const short short_inf = 32767;
const ll less_inf = 0x3f3f3f3f;
const char char_inf = 127;
#pragma GCC optimize(2)
#define accelerate cin.tie(NULL);cout.tie(NULL);ios::sync_with_stdio(false);
#define PI 3.141592653589793
#define EPS 1.0e-8
ll gcd(ll a, ll b) {
    
	return b ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
    
	return a / gcd(a, b) * b;
}
inline ll read() {
    
	ll c = getchar(), Nig = 1, x = 0;
	while (!isdigit(c) && c != '-')c = getchar();
	if (c == '-')Nig = -1, c = getchar();
	while (isdigit(c))x = ((x << 1) + (x << 3)) + (c ^ '0'), c = getchar();
	return Nig * x;
}
inline void out(ll a) {
    
	if (a < 0)putchar('-'), a = -a;
	if (a > 9)out(a / 10);
	putchar(a % 10 + '0');
}
ll qpow(ll x, ll n, ll mod) {
    
	ll res = 1;
	while (n > 0) {
    
		if (n & 1)res = (res * x) % mod;
		x = (x * x) % mod;
		n >>= 1;
	}
	return res;
}
#define read read()
struct node
{
    
	int sta, end, val;
	bool friend operator<(node a, node b)
	{
    
		return a.val < b.val;
	}
}save[10005];
int fa[105];
bool judge[10005];
int mp[105][105];
int find(int x)
{
    
	return fa[x] == x ? x : fa[x] = find(fa[x]);

}
struct link {
    
	int head, end, next;
}Link[105];
void merge(int x, int y)
{
    
	int fx = find(x), fy = find(y);
	if (fx != fy)fa[fx] = fy;
}
int main()
{
    
	int T = read;
	while (T--)
	{
    
		for (int i = 0; i < 105; i++)fa[i] = i;
		memset(judge, 0, sizeof(judge));
		memset(mp, 0, sizeof(mp));
		for (int i = 0; i < 105; i++)
			Link[i].head = Link[i].end = i, Link[i].next = -1;
		int n = read, m = read;
		for (int i = 0; i < m; i++)
			save[i].sta = read, save[i].end = read, save[i].val = read;
		sort(save, save + m);
		int num = 0, cnt = 0;
		int ans = 0;
		for (int i = 0; i < m; i++)
		{
    
			if (find(save[i].sta) != find(save[i].end))
			{
    
				merge(save[i].sta, save[i].end);
				ans += save[i].val;
				judge[i] = true;
				for (int u = Link[save[i].sta].head; ~u; u = Link[u].next)
					for (int v = Link[save[i].end].head; ~v; v = Link[v].next)
						mp[u][v] = mp[v][u] = save[i].val;
				Link[Link[save[i].sta].end].next = Link[save[i].end].head;
				Link[save[i].sta].end = Link[save[i].end].end;
				Link[save[i].end].head = Link[save[i].sta].head;
			}
		}
		int res = int_inf;
		for (int i = 0; i < m; i++)
			if (!judge[i])
				res = min(res, ans - mp[save[i].sta][save[i].end] + save[i].val);
		if (ans == res)puts("Not Unique!");
		else
		{
    
			out(ans);
			puts("");
		}
	}
}

By-Round Moon