study and life share of wss

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 110;
int Map[maxn][maxn];
bool vis[maxn];
int pre[maxn];
int dis[maxn];
int MaxPath[maxn][maxn]; 
//MaxPath[i][j] 指的是在i -> j 在最小生成树的路径上的最大的边
bool used[maxn][maxn];
//判断在最小生成树中是不是有这么一条边
int T, m, n;

void init()
{
    memset(dis, 0x3f, sizeof(dis));
    memset(Map, 0x3f, sizeof(Map));
    memset(vis, 0, sizeof(vis));
    memset(used, 0, sizeof(used));
    memset(pre, 0, sizeof(pre));
}

int primMST()
{
    for(int i = 1; i <= n; i++){
        dis[i] = Map[i][1];
        pre[i] = 1;  //初始化，让每一个点的父亲节点都变成1
    }
    dis[1] = 0; vis[1] = true;
    int ans = 0;

    for(int i = 1; i < n; i++){
        int temp = INF, pos;
        for(int j = 1; j <= n; j++){
            if(!vis[j] && temp > dis[j]){
                temp = dis[j], pos = j;
            }
        }

        if(temp == INF) return -1;
        used[pre[pos]][pos] = used[pos][pre[pos]] = 1;
        ans += dis[pos];
        vis[pos] = true;

        for(int j = 1; j <= n; j++){
            if(vis[j] && j != pos)  MaxPath[j][pos] = MaxPath[pos][j] = max(MaxPath[j][pre[pos]], dis[pos]);
            if(!vis[j] && dis[j] > Map[pos][j]) {dis[j] = Map[pos][j], pre[j] = pos;}
        }
    }
    return ans;
}

int FindSMST(int x)  
//遍历每一条边，让最小生成树加上这一条边，这样就构成了一个回路，然后减掉这两个点之间最大的那一条边
{
    int ans = INF;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            if(i != j && !used[i][j])
                ans = min(ans, x+Map[i][j]-MaxPath[i][j]);
        }
    }
    return ans;
}

int main()
{
    scanf("%d", &T);
    while(T--){
        init();
        scanf("%d %d", &n, &m);
        while(m--){
            int v1, v2, w;
            scanf("%d %d %d", &v1, &v2, &w);
            Map[v1][v2] = Map[v2][v1] = w;
        }
        int ans1 = primMST();
        int ans2 = FindSMST(ans1);
        if(ans1 != ans2) cout << ans1 << endl;
        else             cout << "Not Unique!" << endl;
    }
    return 0;
}