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POJ 2516 最小费用最大流

时间:2014-05-08 13:49:21      阅读:290      评论:0      收藏:0      [点我收藏+]

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每一种货物都是独立的,分成k次最小费用最大流即可!

   1:  /**
   2:  因为e ==0 所以 pe[v]   pe[v]^1 是两条相对应的边
   3:  E[pe[v]].c -= aug;            E[pe[v]^1].c += aug;
   4:  
   5:  */
   6:  #include <queue>
   7:  #include <iostream>
   8:  #include <string.h>
   9:  #include <stdio.h>
  10:  #include <map>
  11:  using namespace std;
  12:  #define V 30010      // vertex
  13:  #define E 150010      // edge
  14:  #define INF 0x3F3F3F3F
  15:  struct MinCostMaxFlow
  16:  {
  17:      struct Edge
  18:      {
  19:          int v, next, cap, cost;  // cap 为容量, cost为单位流量费用
  20:      } edge[E];
  21:   
  22:      int head[V], pe[V], pv[V];            // 每个节点的第一条edge[idx]的编号.
  23:      int  dis[V];            // the shortest path to the src
  24:      bool vis[V];            // visted
  25:      // pe[v]  存放在增广路上到达v的边(u,v) 在edge[]的位置
  26:      // pv[u]  存放在增广路上从u出发的边(u,v) 在edge[]中的位置
  27:      int e, src, sink;            // the index of the edge
  28:      int VertexNum;               // N the num of the vertex. M the num of edges
  29:      void addedge(int  u, int v, int cap, int cost)
  30:      {
  31:          edge[e].v = v, edge[e].cap = cap;
  32:          edge[e].cost = cost,edge[e].next = head[u], head[u] = e++;
  33:          edge[e].v = u, edge[e].cap = 0;
  34:          edge[e].cost = -1*cost, edge[e].next = head[v], head[v] = e++;
  35:      }
  36:      // 求最短路,不存在负环的时候,比 DFS快
  37:      int SPFABFS()
  38:      {
  39:          memset(vis, 0 ,sizeof(vis));
  40:          memset(pv, -1, sizeof(pv));
  41:          for(int i=0; i<V; i++) dis[i] = INF;
  42:          queue<int> Q;
  43:          Q.push(src);
  44:          vis[src] = 1, dis[src] = 0;
  45:          while(!Q.empty())
  46:          {
  47:              int u = Q.front();
  48:              Q.pop();
  49:              vis[u] = 0;
  50:              for(int i=head[u]; i!=-1; i=edge[i].next)
  51:              {
  52:                  int v = edge[i].v;
  53:                  if(edge[i].cap > 0 &&  dis[v] > dis[u] + edge[i].cost  )
  54:                  {
  55:                      dis[v] = dis[u] + edge[i].cost;
  56:                      if(!vis[v])
  57:                      {
  58:                          Q.push(v);
  59:                          vis[v] = 1;
  60:                      }
  61:                      pv[v] = u;
  62:                      pe[v] = i;
  63:                  }
  64:              }
  65:          }
  66:          if(dis[sink] == INF) return -2;          // can‘t from src to sink.
  67:          return dis[sink];
  68:      }
  69:   
  70:      pair<int,int> MCMF()
  71:      {
  72:          int maxflow = 0, mincost = 0;
  73:          while(SPFABFS())
  74:          {
  75:              if(pv[sink] == -1) break;
  76:              int aug = INF;
  77:              for(int i= sink; i!= src; i = pv[i])
  78:                  aug =min(aug, edge[pe[i]].cap);
  79:              maxflow += aug;
  80:              mincost += aug * dis[sink];
  81:              for(int i = sink; i!= src; i = pv[i])
  82:              {
  83:                  edge[pe[i]].cap -= aug;
  84:                  edge[pe[i]^1].cap += aug;
  85:              }
  86:          }
  87:          return make_pair(maxflow, mincost);
  88:      }
  89:      int solve()
  90:      {
  91:          int N,M,K;
  92:          while(scanf("%d%d%d", &N , &M, &K) && N!=0 && M!=0 && K!=0)
  93:          {
  94:              int tmp;
  95:              vector<int> need[55];
  96:              for(int i=0; i<N; i++)
  97:              {
  98:                  for(int j=0; j<K; j++)
  99:                  {
 100:                      scanf("%d", &tmp);
 101:                      need[i].push_back(tmp);
 102:                  }
 103:              }
 104:              vector<int> supply[55];
 105:              for(int i =0; i<M; i++)
 106:              {
 107:                  for(int j=0; j<K; j++)
 108:                  {
 109:                      scanf("%d", &tmp);
 110:                      supply[i].push_back(tmp);
 111:                  }
 112:              }
 113:              vector<vector<vector<int> > > cost;
 114:              for(int i=0; i<K; i++)
 115:              {
 116:                  vector<vector<int> > y;
 117:                  for(int j=0; j<N; j++)
 118:                  {
 119:                      vector<int> x;
 120:                      for(int k = 0; k<M; k++)
 121:                      {
 122:                          scanf("%d", &tmp);
 123:                          x.push_back(tmp);
 124:                      }
 125:                      y.push_back(x);
 126:                  }
 127:                  cost.push_back(y);
 128:              }
 129:              int ret = 0;
 130:              bool flag = true;
 131:              for(int k = 0; k < K; k++)
 132:              {
 133:                  e=0;
 134:                  memset(head, -1,sizeof(head));
 135:                  // M suppliers
 136:                  src = 0;
 137:                  sink = M+N+1;
 138:                  VertexNum = M+N+2;
 139:                  for(int i=0; i<M; i++)
 140:                  {
 141:                      addedge(src,i+1,supply[i][k],0);
 142:                  }
 143:                  int x = 0;
 144:                  for(int i=0; i<N; i++)
 145:                  {
 146:                      x+= need[i][k];
 147:                      addedge(M+i+1,sink, need[i][k], 0);
 148:                  }
 149:                  for(int j=0; j<M; j++)
 150:                  {
 151:                      for(int i=0; i<N; i++)
 152:                      {
 153:                          addedge(j+1, M+i+1,INF,cost[k][i][j]);
 154:                      }
 155:                  }
 156:                  pair<int,int> res = MCMF();
 157:                  if(res.first < x)
 158:                  {
 159:                      flag = 0;
 160:                      break;
 161:                  }
 162:                  ret += res.second;
 163:              }
 164:              if(flag) cout<<ret<<endl;
 165:              else cout<<-1<<endl;
 166:          }
 167:      }
 168:  } mcmf;
 169:   
 170:   
 171:  int main()
 172:  {
 173:      freopen("1.txt","r",stdin);
 174:      mcmf.solve();
 175:      return 0;
 176:  }

POJ 2516 最小费用最大流,布布扣,bubuko.com

POJ 2516 最小费用最大流

标签:style   class   code   ext   color   int   

原文地址:http://www.cnblogs.com/sosi/p/3713917.html

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