网络流基础题目，Edmonds_Karp可解。
1 /* 3549 */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #include <vector> 9 #include <deque> 10 #include <algorithm> 11 #include <cstdio> 12 #include <cmath> 13 #include <ctime> 14 #include <cstring> 15 #include <climits> 16 #include <cctype> 17 #include <cassert> 18 #include <functional> 19 using namespace std; 20 21 #define rep(i, a, n) for (int i=a;i<n;++i) 22 #define per(i, a, n) for (int i=n-1;i>=a;--i) 23 #define pb push_back 24 #define mp make_pair 25 #define all(x) (x).begin(),(x).end() 26 #define SZ(x) ((int)(x).size()) 27 #define lson l, mid, rt<<1 28 #define rson mid+1, r, rt<<1|1 29 30 const int maxn = 20; 31 int F[maxn][maxn]; 32 int P[maxn], a[maxn]; 33 int n, m, ans; 34 int s, t; 35 36 bool bfs() { 37 int u, v; 38 queue<int> Q; 39 40 memset(a, 0, sizeof(a)); 41 a[s] = INT_MAX; 42 Q.push(s); 43 P[s] = s; 44 45 while (!Q.empty()) { 46 u = Q.front(); 47 Q.pop(); 48 for (v=1; v<=n; ++v) { 49 if (!a[v] && F[u][v]) { 50 P[v] = u; 51 Q.push(v); 52 a[v] = min(a[u], F[u][v]); 53 } 54 } 55 } 56 57 return a[t] == 0; 58 } 59 60 void Edmonds_Karp() { 61 int u, v; 62 63 s = 1; 64 t = n; 65 ans = 0; 66 67 while (true) { 68 if (bfs()) 69 break; 70 for (u=t; u!=s; u=P[u]) { 71 F[u][P[u]] += a[t]; 72 F[P[u]][u] -= a[t]; 73 } 74 ans += a[t]; 75 } 76 } 77 78 int main() { 79 int i, j, k; 80 int t, tt; 81 82 #ifndef ONLINE_JUDGE 83 freopen("data.in", "r", stdin); 84 freopen("data.out", "w", stdout); 85 #endif 86 87 scanf("%d", &tt); 88 for (t=1; t<=tt; ++t) { 89 scanf("%d %d", &n, &m); 90 memset(F, 0, sizeof(F)); 91 while (m--) { 92 scanf("%d %d %d", &i, &j, &k); 93 F[i][j] += k; 94 } 95 Edmonds_Karp(); 96 printf("Case %d: %d ", t, ans); 97 } 98 99 #ifndef ONLINE_JUDGE 100 printf("time = %d. ", (int)clock()); 101 #endif 102 103 return 0; 104 }