建分层图，每一层表示一天的情况
从S向第0层的1号点连边，每层的n向T连INF的边
枚举天数，每多一天就多建一层然后跑最大流，如果当前流量大于人数则输出答案
由于路径长度不会超过n，因此tot个人走这条路径总天数不会超过tot + n，故只需要建tot + n层即可
1 /************************************************************** 2 Problem: 1570 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:24 ms 7 Memory:15936 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 const int N = 55; 16 const int M = 5005; 17 const int inf = 1e8; 18 19 inline int read() { 20 int x = 0; 21 char ch = getchar(); 22 while (ch < '0' || '9' < ch) 23 ch = getchar(); 24 while ('0' <= ch && ch <= '9') { 25 x = x * 10 + ch - '0'; 26 ch = getchar(); 27 } 28 return x; 29 } 30 31 struct Edge { 32 int x, y, z; 33 34 inline void read_in() { 35 x = read(), y = read(), z = read(); 36 } 37 } E[M]; 38 39 struct edge { 40 int next, to, f; 41 edge() {} 42 edge(int _n, int _t, int _f) : next(_n), to(_t), f(_f) {} 43 } e[M << 8]; 44 45 int n, m, t; 46 int S, T, ans; 47 int first[M], tot = 1; 48 int d[M], q[M]; 49 50 inline void Add_Edges(int x, int y, int f) { 51 e[++tot] = edge(first[x], y, f), first[x] = tot; 52 e[++tot] = edge(first[y], x, 0), first[y] = tot; 53 } 54 55 #define y e[x].to 56 #define p q[l] 57 bool bfs() { 58 int l, r, x; 59 memset(d, -1, sizeof(d)); 60 d[q[1] = S] = 1; 61 for (l = r = 1; l != r + 1; ++l) 62 for (x = first[p]; x; x = e[x].next) 63 if (!~d[y] && e[x].f) { 64 d[q[++r] = y] = d[p] + 1; 65 if (y == T) return 1; 66 } 67 return 0; 68 } 69 #undef p 70 71 int dfs(int p, int lim) { 72 if (p == T || !lim) return lim; 73 int x, tmp, rest = lim; 74 for (x = first[p]; x && rest; x = e[x].next) 75 if (d[y] == d[p] + 1 && ((tmp = min(e[x].f, rest)) > 0)) { 76 rest -= (tmp = dfs(y, tmp)); 77 e[x].f -= tmp, e[x ^ 1].f += tmp; 78 if (!rest) return lim; 79 } 80 if (rest) d[p] = -1; 81 return lim - rest; 82 } 83 #undef y 84 85 int Dinic() { 86 int res = 0; 87 while (bfs()) 88 res += dfs(S, inf); 89 return res; 90 } 91 92 int main() { 93 int i, j; 94 n = read(), m = read(), t = read(); 95 for (i = 1; i <= m; ++i) E[i].read_in(); 96 S = 1, T = 2; 97 Add_Edges(S, 3, t); 98 for (i = 1; i <= n + t; ++i) { 99 for (j = 1; j <= m; ++j) 100 Add_Edges(i * n - n + E[j].x + 2, i * n + E[j].y + 2, E[j].z); 101 for (j = 1; j <= n; ++j) 102 Add_Edges(i * n - n + j + 2, i * n + j + 2, inf); 103 Add_Edges(i * n + n + 2, T, inf); 104 ans += Dinic(); 105 if (ans == t) { 106 printf("%d ", i); 107 return 0; 108 } 109 } 110 }