BZOJ1570 [JSOI2008]Blue Mary的旅行

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

 /**************************************************************
     Problem: 1570
     User: rausen
     Language: C++
     Result: Accepted
     Time:24 ms
     Memory:15936 kb
 ****************************************************************/

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;
 const int N = ;
 const int M = ;
 const int inf = 1e8;

 inline int read() {
     int x = ;
     char ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }

 struct Edge {
     int x, y, z;

     inline void read_in() {
         x = read(), y = read(), z = read();
     }
 } E[M];

 struct edge {
     int next, to, f;
     edge() {}
     edge(int _n, int _t, int _f) : next(_n), to(_t), f(_f) {}
 } e[M << ];

 int n, m, t;
 int S, T, ans;
 int first[M], tot = ;
 int d[M], q[M];

 inline void Add_Edges(int x, int y, int f) {
     e[++tot] = edge(first[x], y, f), first[x] = tot;
     e[++tot] = edge(first[y], x, ), first[y] = tot;
 }

 #define y e[x].to
 #define p q[l]
 bool bfs() {
   int l, r, x;
   memset(d, -, sizeof(d));
   d[q[] = S] = ;
   for (l = r = ; l != r + ; ++l)
     for (x = first[p]; x; x = e[x].next)
       if (!~d[y] && e[x].f) {
     d[q[++r] = y] = d[p] + ;
     if (y == T) return ;
       }
   return ;
 }
 #undef p

 int dfs(int p, int lim) {
   if (p == T || !lim) return lim;
   int x, tmp, rest = lim;
   for (x = first[p]; x && rest; x = e[x].next)
     if (d[y] == d[p] +  && ((tmp = min(e[x].f, rest)) > )) {
       rest -= (tmp = dfs(y, tmp));
       e[x].f -= tmp, e[x ^ ].f += tmp;
       if (!rest) return lim;
     }
   if (rest) d[p] = -;
   return lim - rest;
 }
 #undef y

 int Dinic() {
   int res = ;
   while (bfs())
     res += dfs(S, inf);
   return res;
 }

 int main() {
     int i, j;
     n = read(), m = read(), t = read();
     for (i = ; i <= m; ++i) E[i].read_in();
     S = , T = ;
     Add_Edges(S, , t);
     for (i = ; i <= n + t; ++i) {
         for (j = ; j <= m; ++j)
             Add_Edges(i * n - n + E[j].x + , i * n + E[j].y + , E[j].z);
         for (j = ; j <= n; ++j)
             Add_Edges(i * n - n + j + , i * n + j + , inf);
         Add_Edges(i * n + n + , T, inf);
         ans += Dinic();
         if (ans == t) {
             printf("%d\n", i);
             return ;
         }
     }
 }