Poj 2516 Minimum Cost (min. maximum flow)

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Poj 2516 Minimum Cost

Title Description:

There are n stores, M warehouses, and each store and warehouse has K-type goods. Why Now N stores are starting to send orders to M warehouses, and the order information is the current store demand for each type of goods. Different stores from different warehouses to buy different goods cost different, ask is not able to meet all the requirements of the store, if the total cost of the store to the minimum amount?

Problem Solving Ideas:

Simple cost flow, to run K times the minimum cost of the maximum flow, each time only a kind of goods to build a chart run cost flow. Each build takes 0 as the source point, [1,m] is a warehouse, [M+1, N+m] is a store, and N+m+1 is a meeting point. 0 with [1,m] The edge, the edge capacity for the warehouse inventory, the cost of the side unit traffic is 0.

[1,m] with [m+1,n+m] edge, edge capacity for INF, the cost of the side unit traffic for the corresponding warehouse supply I goods to the unit of the corresponding store cost. [M+1,n+m] with the sink point edge, the edge capacity for the store demand, the cost of the side unit traffic is 0. The topic of the map when the initialization must be done well, or it is easy to tle.

1#include <queue>2#include <cstdio>3#include <cstring>4#include <iostream>5#include <algorithm>6 using namespacestd;7 8 Const intMAXN = About;9 Const intINF =0x3f3f3f3f;Ten intFLOW[MAXN][MAXN], CAP[MAXN][MAXN]; One intSHOP[MAXN][MAXN], SUPPLY[MAXN][MAXN]; A intVIS[MAXN], PRE[MAXN], DIS[MAXN], COST[MAXN][MAXN]; - intC, Flow, S, E; -  the BOOLSPFA () - { -Queue <int>Q; -      for(intI=s; i<=e; i++) +     { -Dis[i] =INF; +Vis[i] =0; A     } atdis[0] =0; -vis[0] =1; - Q.push (s); -      while(!q.empty ()) -     { -         intU =Q.front (); in Q.pop (); -Vis[u] =0; to          for(intV=s; v<=e; v++) +         { -             if(Cap[u][v]>flow[u][v] && dis[v]>dis[u]+Cost[u][v]) the             { *DIS[V] = Dis[u] +Cost[u][v]; $PRE[V] =u;Panax Notoginseng                 if(!Vis[v]) -                 { theVIS[V] =1; + Q.push (v); A                 } the             } +         } -     } $     if(Dis[e] = =INF) $         return false; -     return true; - } the  - voidMincostmaxflow ()Wuyi { thememset (Flow,0,sizeof(flow)); -c = Flow =0; Wu      while(SPFA ()) -     { About         intMin =INF; $          for(intI=e; I!=s; I=Pre[i]) -min = min (min, cap[pre[i]][i]-flow[pre[i]][i]); -          for(intI=e; I!=s; I=Pre[i]) -         { AFlow[pre[i]][i] + =Min; +Flow[i][pre[i]]-=Min; the         } -Flow + =Min; $c + = Min *Dis[e]; the     } the } the  the intMain () - { in     intN, M, K; the      while(SCANF (" %d%d%d", &n, &m, &k), n| | m| |k) the     { About         intflag, cost; theFlag = Cost = s =0, E = n + M +1; the          the          for(intI=1; i<=n; i++) +              for(intj=1; j<=k; J + +) -scanf ("%d", &shop[i][j]); the Bayi          for(intI=1; i<=m; i++) the              for(intj=1; j<=k; J + +) thescanf ("%d", &supply[i][j]); -                  -memset (Cap,0,sizeof(CAP)); thememset (Cost,0,sizeof(cost)); the          for(intI=1; i<=m; i++) the              for(inti=n+1; j<=n+m; J + +) theCAP[I][J] =INF; -  the          for(intI=1; i<=k; i++) the         { the             intTotal =0;94              for(intj=1; j<=n; J + +) the                  for(intL=1; l<=m; l++) the                 { thescanf ("%d", &cost[l][j+m]);98COST[J+M][L] =-cost[l][j+m]; About                 } -                 101             if(flag)102                 Continue;103 104              for(intj=1; j<=n; J + +) the                 {106Cap[j+m][e] =Shop[j][i];107Total + =Shop[j][i];108                 }109              for(intj=1; j<=m; J + +) theCAP[S][J] =Supply[j][i];111  the Mincostmaxflow ();113             if(Flow <Total ) theFlag =1; theCost + =C; the             117         }118printf ("%d\n", flag?-1: Cost);119     } -     return 0;121}

Poj 2516 Minimum Cost (min. maximum flow)