思路：

欧拉图

定理：一个度数为奇数的点的个数小于等于2的联通图存在欧拉回路

对于这道题目的图，点的个数为4，所以最坏的情况下4个点的度数都为奇数，在这种情况下只要删去一条边就可以满足条件了

欧拉回路算法：大圈小圈法，从起点开始跑每条边，把每条遍标记一下，直到跑到某个位置不能跑了，把点如栈，最后倒着输出

所以枚举删掉的边，跑联通图，最后判断联通图是否符合条件，复杂度：O(n^2)

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
     
      using namespace std;#define fi first#define se second#define pi acos(-1.0)#define LL long long#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
      
       #define pii pair
       
        #define piii pair
        
         #define mem(a, b) memset(a, b, sizeof(a))#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);//headconst int N = 120;struct edge {    int to, w, id;};vector
         
           g[N]; int d[5];pii a[N];bool vis[N], node[5];LL ans, tot = 0;void dfs(int u) {    node[u] = true;    for (int i = 0; i < g[u].size(); i++) {        int id = g[u][i].id;        if(!vis[id]) {            vis[id] = true;            dfs(g[u][i].to);            tot += g[u][i].w;        }    }}int main() {    int n, u, v, w;    scanf("%d", &n);    for (int i = 1; i <= n; i++) {        scanf("%d %d %d", &u, &w, &v);        g[u].pb(edge{v, w, i});        g[v].pb(edge{u, w, i});        d[u]++;        d[v]++;        a[i].fi = u;        a[i].se = v;    }    ans = 0;    for (int i = 0; i <= n; i++) {        if(i != 0 && a[i].fi == a[i].se) continue;         for (int j = 0; j <= n; j++) vis[j] = false;        vis[i] = true;        d[a[i].fi] --;        d[a[i].se] --;        for (int j = 1; j <= 4; j++) {            tot = 0;            mem(node, false);            dfs(j);            int cnt = 0;            for (int k = 1; k <= 4; k++) if(node[k] && (d[k]&1)) cnt++;            if(cnt <= 2)ans = max(ans, tot);        }        d[a[i].fi]++;        d[a[i].se]++;     }    printf("%lld\n", ans);    return 0;}