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.

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!

次小生成树求一波，判断其总权值是否与最小生成树总权值一致，不一致就Unique，一致就Not Unique!。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 111;
const int inf = 0x3f3f3f3f;
int Map[maxn][maxn];//邻接矩阵存图
int Max[maxn][maxn];//表示最小生成树中i到j的最大边权
bool used[maxn][maxn];//判断该边是否加入最小生成树
int pre[maxn];
int dis[maxn];
void init(int n)
{
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (i == j) Map[i][j] = 0;
            else Map[i][j] = inf;
}
void read(int m)
{
    int u, v, w;
    for (int i = 0; i < m; i++)
    {
        scanf("%d %d %d", &u, &v, &w);
        Map[u][v] = Map[v][u] = w;
    }
}
int prim(int n)
{
    int ans = 0;
    bool vis[maxn];
    memset(vis, false, sizeof(vis));
    memset(used, false, sizeof(used));
    memset(Max, 0, sizeof(Max));
    for (int i = 2; i <= n; i++) 
    {
        dis[i] = Map[1][i];
        pre[i] = 1;
    }
    pre[1] = 0;
    dis[1] = 0;
    vis[1] = true;
    for (int i = 2; i <= n; i++)
    {
        int min_dis = inf, k;
        for (int j = 1; j <= n; j++)
        {
            if (!vis[j] && min_dis > dis[j])
            {
                min_dis = dis[j];
                k = j;
            }
        }
        if (min_dis == inf) return -1;//如果不存在最小生成树
        ans += min_dis;
        vis[k] = true;
        used[k][pre[k]] = used[pre[k]][k] = true;
        for (int j = 1; j <= n; j++)
        {
            if (vis[j]) Max[j][k] = Max[k][j] = max(Max[j][pre[k]], dis[k]);
            if (!vis[j] && dis[j] > Map[k][j])
            {
                dis[j] = Map[k][j];
                pre[j] = k;
            }
        }
    }
    return ans;//最小生成树的权值之和
}
int smst(int n, int min_ans)//min_ans 是最小生成树的权值和
{
    int ans = inf;
    for (int i = 1; i <= n; i++)//枚举最小生成树之外的边
        for (int j = i + 1; j <= n; j++)
            if (Map[i][j] != inf && !used[i][j])
                ans = min(ans, min_ans + Map[i][j] - Max[i][j]);
    if (ans == inf) return -1;
    return ans;
}
void solve(int n)
{
    int ans = prim(n);
    if (ans == -1)
    {
        puts("Not Unique!");
        return;
    }
    if (smst(n, ans) == ans)
    printf("Not Unique!\n");
    else
    printf("%d\n", ans);
}

int main()
{
    int T, n, m;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d %d", &n, &m);
        init(n);
        read(m);
        solve(n);
    }
    return 0;
}