ZOJ 2314 带上下界的可行流

对于无源汇问题,方法有两种.

1 从边的角度来处理. 新建超级源汇, 对于每一条有下界的边,x->y, 建立有向边 超级源->y ,容量为x->y下界,建立有向边 x-> 超级汇,容量为x->y下界.建立有向边 x->y,容量为x->y的上界减下界.

2 从点的角度来处理. 新建超级源汇,对于每个点流进的下界和为 in, 流出此点的下界和为out.如果in > out. 建立有向边 超级源->i,容量为in-out.反之,建立有向边 i->超级汇,容量为out-in.

如果超级源的每一条出边都满流,则存在一个可行流,可行流的流量就是每一条逆向边的流量+ 此边下界.

从点的角度的处理方法是边的处理方法的拓展.基于边的处理比较容易理解,但是复杂度较高.

下面是基于点的处理的程序:

   1:  /*

   2:      带上下界的可行流

   3:  */

   4:  #include<iostream>

   5:  #include<cmath>

   6:  #include<memory>

   7:  #include <string.h>

   8:  #include <cstdio>

   9:  using namespace std;

  10:   

  11:  #define V 300      // vertex

  12:  #define E  40800     // edge

  13:  #define INF 0x3F3F3F3F  // 1061109567

  14:   

  15:  int i,j,k;

  16:  #define REP(i,n) for((i)=0;(i)<(int)(n);(i)++)

  17:  #define snuke(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)

  18:   

  19:  struct MaxFlow

  20:  {

  21:      struct Edge

  22:      {

  23:          int v, w, next;     //w for capicity

  24:          int lb,up;

  25:      } edge[E];

  26:   

  27:      int head[V];          // head[u]表示顶点u第一条邻接边的序号, 若head[u] = -1, u没有邻接边

  28:      int e;                // the index of the edge

  29:      int src, sink;

  30:      int net[V];              // 流入此节点的流的下界和 - 流出此节点的流的下界和,对于带上下界的来进行使用

  31:   

  32:   

  33:      void addedge(int  u, int v, int w, int lb = 0, int up = INF, int rw = 0)

  34:      {

  35:          edge[e].v = v;

  36:          edge[e].w= w;

  37:          edge[e].next = head[u];

  38:          edge[e].lb = lb, edge[e].up = up;

  39:          head[u] = e++;

  40:          // reverse edge  v -> u

  41:          edge[e].v = u;

  42:          edge[e].w = rw;

  43:          edge[e].lb = lb, edge[e].up = up;

  44:          edge[e].next = head[v];

  45:          head[v] = e++;

  46:      }

  47:   

  48:      int ISAP(int VertexNum )

  49:      {

  50:          int u, v, max_flow, aug, min_lev;

  51:          int curedge[V], parent[V], level[V];

  52:          int count[V], augment[V];

  53:   

  54:          memset(level, 0, sizeof(level));

  55:          memset(count, 0, sizeof(count));

  56:          REP(i,VertexNum+1) curedge[i] = head[i];

  57:          max_flow = 0;

  58:          augment[src] = INF;

  59:          parent[src] = -1;

  60:          u = src;

  61:   

  62:          while (level[src] < VertexNum)

  63:          {

  64:              if (u == sink)

  65:              {

  66:                  max_flow += augment[sink];

  67:                  aug = augment[sink];

  68:                  for (v = parent[sink]; v != -1; v = parent[v])

  69:                  {

  70:                      i = curedge[v];

  71:                      edge[i].w  -= aug;

  72:                      edge[i^1].w  += aug;

  73:                      augment[edge[i].v] -= aug;

  74:                      if (edge[i].w == 0) u = v;

  75:                  }

  76:              }

  77:              for (i = curedge[u]; i != -1; i = edge[i].next)

  78:              {

  79:                  v = edge[i].v;

  80:                  if (edge[i].w > 0 && level[u] == (level[v]+1))

  81:                  {

  82:                      augment[v] = min(augment[u], edge[i].w);

  83:                      curedge[u] = i;

  84:                      parent[v] = u;

  85:                      u = v;

  86:                      break;

  87:                  }

  88:              }

  89:              if (i == -1)

  90:              {

  91:                  if (--count[level[u]] == 0) break;

  92:                  curedge[u] = head[u];

  93:                  min_lev = VertexNum;

  94:                  for (i = head[u]; i != -1; i = edge[i].next)

  95:                      if (edge[i].w > 0)

  96:                          min_lev = min(level[edge[i].v], min_lev);

  97:                  level[u] = min_lev + 1;

  98:                  count[level[u]]++;

  99:                  if (u != src ) u = parent[u];

 100:              }

 101:          }

 102:          return max_flow;

 103:      }

 104:      void solve()

 105:      {

 106:          int T; cin>>T;

 107:          for(int cas =1; cas <=T; cas++)

 108:          {

 109:              int N,M; cin>>N>>M;

 110:              e = 0;

 111:              memset(head, -1, sizeof(head));

 112:              memset(net, 0, sizeof(net));

 113:              int a,b,c,d;

 114:              for(int i=1; i<=M;i++)

 115:              {

 116:                  scanf("%d%d%d%d", &a, &b, &c, &d);

 117:                  net[b] += c; net[a] -= c;

 118:                  addedge(a,b,d-c, c,d);

 119:              }

 120:              // 无源汇,这两个是super src, super sink.

 121:              src =N+1; sink = N+2;

 122:              for(int i=1; i<=N; i++)

 123:              {

 124:                  if(net[i] > 0)

 125:                      addedge(src, i, net[i]);

 126:                  else

 127:                      addedge(i, sink, -net[i]);

 128:              }

 129:              ISAP(N+2);

 130:              bool flag = true;

 131:              for (i = head[src]; i != -1; i = edge[i].next)

 132:              {

 133:                  if(edge[i].w !=0)

 134:                  {

 135:                      flag = false;

 136:                      break;

 137:                  }

 138:              }

 139:              if(flag)

 140:              {

 141:                  cout<<"YES"<<endl;

 142:                  for(int i=0; i<M;i++)

 143:                  {

 144:                      cout<<edge[i*2+1].w + edge[i*2+1].lb<<endl;

 145:                  }

 146:                  cout<<endl;

 147:              }else

 148:              {

 149:                  cout<<"NO"<<endl;

 150:                  cout<<endl;

 151:              }

 152:          }

 153:   

 154:      }

 155:  }sap;

 156:   

 157:  int main()

 158:  {

 159:  //    freopen("1.txt","r",stdin);

 160:      sap.solve();

 161:      return 0;

 162:  }

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