Code 1035 train stop [maximum fee flow]

Question: Code 1035 train stop

Question: Chinese Question. Read it directly.

Analysis: after reading the question, it is found that the maximum cost is not easy to take into account the cost flow. In fact, the maximum and minimum costs are the same. If the maximum cost is, the minimum cost can be changed to a negative value.

Graph creation method:

Super source point SS connection s, capacity is N, cost is 0

Split each train into two nodes I and II. The Edge building capacity is 1, and the cost is the opposite of the money paid to the station.

S connection I, capacity INF, cost 0

II connection T, capacity INF, cost 0

Enumeration edge creation before the last point of all time points. The capacity is 1 and the cost is 0.

And then calculate a billing flow. The opposite number is ans.

AC code:

# Include <cstdio> # include <iostream> # include <algorithm> # include <vector> # include <cstring> # include <queue> # include <string> # include <map> using namespace STD; # define del (a, B) memset (a, B, sizeof (A) const int INF = 0x3f3f3f3f; const int n = 220; struct node {int from,, cap, flow; double cost ;}; vector <node> E; vector <int> V [N]; int vis [N]; double dis [N]; int P [N], a [n]; // P save father, a save capvoid clear (int x) {fo R (INT I = 0; I <= x; I ++) V [I]. clear (); E. clear ();} void add_node (int from, int to, int cap, double cost) // Shab {e. push_back (node) {from, to, Cap, 0, cost}); E. push_back (node) {to, from, 0, 0,-cost}); int Len = E. size ()-1; V [to]. push_back (LEN); V [from]. push_back (len-1);} bool bellmanford (int s, int T, Int & flow, double & cost) {for (INT I = 0; I <= N; I ++) dis [I] = 1000000000; del (VIS, 0); DIS [s] = 0; vis [s] = 1; p [s] = 0; [S] = inf; queue <int> q; q. Push (s); While (! Q. empty () {int u = Q. front (); q. pop (); vis [u] = 0; For (INT I = 0; I <V [u]. size (); I ++) {node & G = E [V [u] [I]; If (G. cap> G. flow & dis [G. to]> dis [u] + G. cost) {dis [G. to] = dis [u] + G. cost; // printf ("%. 5lf \ n ", G. cost); P [G. to] = V [u] [I]; // Save the precursor A [G. to] = min (A [u], G. cap-g.flow); If (! Vis [G. to]) {q. push (G. to); vis [G. to] = 1 ;}}}if (DIS [T] = 1000000000) return false; flow + = A [T]; cost + = dis [T] * A [T]; int u = T; while (u! = S) {e [p [u]. flow + = A [T]; E [p [u] ^ 1]. flow-= A [T]; u = E [p [u]. from;} return true;} double min_cost (int s, int t) {int flow = 0; double cost = 0; while (bellmanford (S, T, flow, cost); Return cost;} struct tree {int fir, SEC; double val;}; tree KF [200]; int main () {int n, m; while (~ Scanf ("% d", & N, & M) {int Ss = 0, S = 1; int t = 2 * m + 3, tt = t + 1; for (INT I = 1; I <= m; I ++) {scanf ("% d % lf % d", & KF [I]. FIR, & KF [I]. val, & KF [I]. sec); KF [I]. SEC + = KF [I]. FIR; add_node (S, 2 * I, INF, 0); double cs = KF [I]. val * 0.01; // printf ("%. 5lf \ n ", CS); add_node (2 * I, 2 * I + 1,1,-CS); add_node (2 * I + 1, t, INF, 0 );} for (INT I = 1; I <= m; I ++) {for (Int J = 1; j <= m; j ++) // note {if (I = J) continue; If (kf [J]. fir> KF [I]. sec) add_node (I * 2 + 1, J * 0.0, 0) ;}} add_node (SS, S, N,); double ans = min_cost (SS, t ); printf ("%. 2lf \ n ",-ans); clear (TT);} return 0 ;}

Code 1035 train stop [maximum fee flow]