1 /*若原圖的基圖不連通, 2 或者存在某個點的入度或出度為 0 則無解。 3 統計所有點的入度出度之差 Di, 對于 Di > 0 的點, 4 加邊(s, i, Di, 0); 對于 Di < 0 的點加邊(i, t, -Di,0); 5 對原圖中的每條邊(i, j), 6 在網絡中加邊(i, j, ∞, Dij),Dij 為邊(i, j)的權值。 7 求一次最小費用流,費用加上原圖所有邊權和即為結果。 8 若進一步要求輸出最小權值回路,則對所有流量 fij > 0 的邊(i, j),在原圖中復制fij 份,這樣原圖便成為歐拉圖,求一次歐拉回路即可。 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 }
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