Poj optimal milking

Optimal milking

 

Question:

There are K machines and C is a cow. It is required to find the shortest distance from the nearest cow to each milk generator. A matrix of C + k is given, indicating the distance between various labels.

Each place has a maximum of M cattle.

 

Algorithm analysis:

Binary + Shortest Path + network stream

Difficult to think about. I am reading the solution report. Then, I reached my hand. Wrong started three times. After that, an unintentional change would be AC. Silent ....

When wrong is in progress, the Network Flow Graph can only be the distance between the machine and the cow. The cows and cows or machines do not need edge construction. At that time, my brains were exhausted !!!!

# Include <iostream> # include <algorithm> # include <vector> # include <queue> # include <cstdio> # include <cstring> using namespace STD; const int INF = 1 <20; const int maxn = 1000; struct edge {int from, to, Cap, flow, cost; edge (){}; edge (INT _ from, int _ to, int _ cap, int _ flow): From (_ from), to (_ to), Cap (_ cap ), flow (_ flow) {};}; vector <edge> edges; vector <int> G [maxn]; int cur [maxn], d [maxn]; bool VST [maxn]; int SRC, sink; Int Dist [maxn] [maxn]; int K, C, M, V; void Init () {src = V + 1; sink = SRC + 1; for (INT I = 0; I <= sink; ++ I) g [I]. clear (); edges. clear () ;}void flody () {for (int K = 0; k <v; ++ K) for (INT I = 0; I <v; ++ I) for (Int J = 0; j <v; ++ J) if (Dist [I] [J]> Dist [I] [k] + dist [k] [J]) dist [I] [J] = DIST [I] [k] + dist [k] [J]; // For (INT I = 0; I <V; ++ I) {// For (Int J = 0; j <v; ++ J) // printf ("% d", DIST [I] [J]); // puts ("");//}} Void addedge (int from, int to, int cap) {edges. push_back (edge (from, to, Cap, 0); edges. push_back (edge (to, from, 0, 0); int SZ = edges. size (); G [from]. push_back (SZ-2); G [to]. push_back (SZ-1);} void build (INT limit) {Init (); For (INT I = K; I <v; ++ I) {// cow and Source Vertex addedge (SRC, I, 1) ;}for (INT I = 0; I <K; ++ I) {// machine and sink addedge (I, sink, m);} // note ---> I = K !!! J <K !!!! For (INT I = K; I <v; ++ I) {// connection between the cow and the machine for (Int J = 0; j <K; ++ J) {If (Dist [I] [J] <= Limit) {addedge (I, j, 1) ;}}} bool BFS () {memset (VST, 0, sizeof (VST); queue <int> q; q. push (SRC); D [SRC] = 0; VST [SRC] = 1; while (! Q. empty () {int x = Q. front (); q. pop (); For (INT I = 0; I <(INT) g [X]. size (); ++ I) {edge & E = edges [G [x] [I]; If (! VST [E. to] & E. cap> E. flow) {VST [E. to] = 1; d [E. to] = d [x] + 1; q. push (E. to) ;}}return VST [sink];} int DFS (int x, int A) {If (x = sink | A = 0) return; int flow = 0, F; For (Int & I = cur [X]; I <(INT) g [X]. size (); ++ I) {edge & E = edges [G [x] [I]; If (d [E. to] = d [x] + 1 & (F = DFS (E. to, min (A, E. cap-e. flow)> 0) {e. flow + = f; edges [G [x] [I] ^ 1]. flow-= f; flow + = f; A-= f; if (a = 0) break;} retur N flow;} int maxflow () {int flow = 0; while (BFS () {memset (cur, 0, sizeof (cur); flow + = DFS (SRC, INF);} return flow;} bool check (INT mid) {build (MID); int flow = maxflow (); // cout <"Flow: "<flow <Endl; return flow = C;} void solve () {flody (); int lB =-1, UB = inf + 100; while (UB-lb> 1) {int mid = (Lb + UB)/2; If (check (MID) UB = mid; else lB = mid; // cout <"mid:" <mid <"LB:" <LB <"UB:" <UB <Endl;} printf ("% d \ n", UB);} int main () {// freopen ("input.txt", "r", stdin); While (~ Scanf ("% d", & K, & C, & M) {v = K + C; int X; For (INT I = 0; I <v; ++ I) {for (Int J = 0; j <v; ++ J) {scanf ("% d", & X ); dist [I] [J] = (x = 0? INF: X);} Dist [I] [I] = 0;} solve ();} return 0 ;}

 

Another method is multi-match, which has not been written yet. Complete it.

Poj optimal milking