1 /*若原图的基图不连通,
2 或者存在某个点的入度或出度为 0 则无解。
3 统计所有点的入度出度之差 Di, 对于 Di > 0 的点,
4 加边(s, i, Di, 0); 对于 Di < 0 的点加边(i, t, -Di,0);
5 对原图中的每条边(i, j),
6 在网络中加边(i, j, ∞, Dij),Dij 为边(i, j)的权值。
7 求一次最小费用流,费用加上原图所有边权和即为结果。
8 若进一步要求输出最小权值回路,则对所有流量 fij > 0 的边(i, j),在原图中复制fij 份,这样原图便成为欧拉图,求一次欧拉回路即可。
9 */
10 #include <iostream>
11 #include <cstdio>
12 #include <cstring>
13 #include <queue>
14 #include <algorithm>
15 #include <cmath>
16
17 using namespace std;
18
19 const int maxn = 1e2 + 5;
20 const int maxm = 2e4 + 5;
21 const int inf = 0x3f3f3f3f;
22
23 struct MCMF {
24 struct Edge {
25 int v, c, w, next;
26 }p[maxm << 1];
27 int e, head[maxn], dis[maxn], pre[maxn], cnt[maxn], sumFlow, n;
28 bool vis[maxn];
29 void init(int nt){
30 e = 0, n = nt + 1;
31 memset(head, -1, sizeof(head[0]) * (n + 2));
32 }
33 void addEdge(int u, int v, int c, int w){
34 p[e].v = v; p[e].c = c; p[e].w = w; p[e].next = head[u]; head[u] = e++;
35 swap(u, v);
36 p[e].v = v; p[e].c = 0; p[e].w = -w; p[e].next = head[u]; head[u] = e++;
37 }
38 bool spfa(int S, int T){
39 queue <int> q;
40 for (int i = 0; i <= n; ++i)
41 vis[i] = cnt[i] = 0, pre[i] = -1, dis[i] = inf;
42 vis[S] = 1; dis[S] = 0;
43 q.push(S);
44 while (!q.empty()){
45 int u = q.front(); q.pop();
46 vis[u] = 0;
47 for (int i = head[u]; i + 1; i = p[i].next){
48 int v = p[i].v;
49 if (p[i].c && dis[v] > dis[u] + p[i].w){
50 dis[v] = dis[u] + p[i].w;
51 pre[v] = i;
52 if (!vis[v]){
53 q.push(v);
54 vis[v] = 1;
55 if (++cnt[v] > n) return 0;
56 }
57 }
58 }
59 }
60 return dis[T] != inf;
61 }
62 int mcmf(int S, int T){
63 sumFlow = 0;
64 int minFlow = 0, minCost = 0;
65 while (spfa(S, T)){
66 minFlow = inf + 1;
67 for (int i = pre[T]; i + 1; i = pre[ p[i ^ 1].v ])
68 minFlow = min(minFlow, p[i].c);
69 sumFlow += minFlow;
70 for (int i = pre[T]; i + 1; i = pre[ p[i ^ 1].v ]){
71 p[i].c -= minFlow;
72 p[i ^ 1].c += minFlow;
73 }
74 minCost += dis[T] * minFlow;
75 }
76 return minCost;
77 }
78 int ind[maxn], outd[maxn], ans ;
79 bool build(int nt, int mt){
80 init(nt);
81 memset(ind, 0, sizeof(ind));
82 memset(outd, 0, sizeof(outd));
83 ans = 0;
84 int u, v, c;
85 while (mt--){
86 scanf("%d%d%d", &u, &v, &c);
87 u++, v++;
88 addEdge(u, v, inf, c);
89 ans += c;
90 outd[u]++, ind[v]++;
91 }
92 for (int i = 1; i <= nt; ++i){
93 if (ind[i] == 0 || outd[i] == 0) return false;
94 }
95 for (int i = 1; i <= nt; ++i){
96 if (ind[i] - outd[i] > 0)
97 addEdge(0, i, ind[i] - outd[i], 0);
98 else if (ind[i] - outd[i] < 0)
99 addEdge(i, n, outd[i] - ind[i], 0);
100 }
101 return true;
102 }
103 void solve(){
104 ans += mcmf(0, n);
105 printf("%d\n", ans);
106 }
107 }my;
108 int main(){
109 int tcase, n, m;
110 scanf("%d", &tcase);
111 while (tcase--){
112 scanf("%d%d", &n, &m);
113 if (!my.build(n, m)){
114 printf("-1\n");
115 continue;
116 }
117 my.solve();
118 }
119 return 0;
120 }
