[24 PowerOJ1752 & network flow problem] transportation issues (cost flow)

Meaning of the questions:

 

 

 Ideas:

【problem analysis】

Cost flow problem.

[Modeling]

Warehouse seen all vertices in the bipartite graph Xi, all retailers seen bipartite graph vertices Yi, T. sink establishing additional source S

1, Xi is connected to each of S from a number of goods in the warehouse capacity of AI, 0 cost directed edge.

2, in order to store the required quantity of goods from each Yi bi T to connect a capacity at a cost of zero directed edges.

3, connected to each from each Xi Yj an infinite capacity, cost cij directed edge.

Seeking a minimum cost maximum flow, minimum cost flow value is the minimum shipping, seeking the most cost maximum flow, maximum flow value is the most cost shipping.

[Modeling Analysis]

Each warehouse to imagine a transit point by point source ai units shipped cargo, freight is zero, each store also as a transit station, transport to the target meeting point bi cargo units. There is a road between each warehouse and retail stores, the capacity is infinite, the unit cost of freight cij. Request from the source

To sink the cost flow, is shipping.

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 typedef unsigned int uint;
  5 typedef unsigned long long ull;
  6 typedef long double ld;
  7 typedef pair<int,int> PII;
  8 typedef pair<ll,ll> Pll;
  9 typedef vector<int> VI;
 10 typedef vector<PII> VII;
 11 typedef pair<ll,ll>P;
 12 #define N  100010
 13 #define M  1000000
 14 #define INF 1e9
 15 #define fi first
 16 #define se second
 17 #define MP make_pair
 18 #define pb push_back
 19 #define pi acos(-1)
 20 #define mem(a,b) memset(a,b,sizeof(a))
 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 23 #define lowbit(x) x&(-x)
 24 #define Rand (rand()*(1<<16)+rand())
 25 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 26 #define ls p<<1
 27 #define rs p<<1|1
 28 
 29 const ll MOD=1e9+7,inv2=(MOD+1)/2;
 30       double eps=1e-6;
 31       int dx[4]={-1,1,0,0};
 32       int dy[4]={0,0,-1,1};
 33 
 34 int head[N],vet[N],len1[N],len2[N],nxt[N],dis[N],q[N],inq[N],a[N],b[N],c[500][500],
 35     pre[N][2],s,S,T,tot,ans1,ans2,n,m;
 36 
 37 
 38 
 39 int read()
 40 {
 41    int v=0,f=1;
 42    char c=getchar();
 43    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 44    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 45    return v*f;
 46 }
 47 
 48 void add(int a,int b,int c,int d)
 49 {
 50     nxt[++tot]=head[a];
 51     vet[tot]=b;
 52     len1[tot]=c;
 53     len2[tot]=d;
 54     head[a]=tot;
 55 
 56     nxt[++tot]=head[b];
 57     vet[tot]=a;
 58     len1[tot]=0;
 59     len2[tot]=-d;
 60     head[b]=tot;
 61 }
 62 
 63 void addedge()
 64 {
 65     tot=1;
 66     rep(i,1,s) head[i]=0;
 67     rep(i,1,n) add(S,i,a[i],0);
 68     rep(i,1,m) add(i+n,T,b[i],0);
 69     rep(i,1,n)
 70      rep(j,1,m) add(i,j+n,INF,c[i][j]);
 71 }
 72 
 73 int spfa1()
 74 {
 75     rep(i,1,s)
 76     {
 77         dis[i]=INF;
 78         inq[i]=0;
 79     }
 80     int t=0,w=1;
 81     q[1]=S; dis[S]=0; inq[S]=1;
 82     while(t<w)
 83     {
 84         t++; int u=q[t%(s+5)]; inq[u]=0;
 85         int e=head[u];
 86         while(e)
 87         {
 88             int v=vet[e];
 89             if(len1[e]&&dis[u]+len2[e]<dis[v])
 90             {
 91                 dis[v]=dis[u]+len2[e];
 92                 pre [v] [ 0 ] = u;
93                  pre [v] [ 1 ] = e;
94                  if (! Inq [v])
 95                  {
 96                      w ++; q [w% (s + 5 )] = v; inq [v] = 1 ;
97                  }
 98              }
 99              e = nxt [e];
100          }
 101      }
 102      if (dis [T] == INF) return  0 ;
103      return  1 ;
104  }
105 
106 int spfa2()
107 {
108     rep(i,1,s)
109     {
110         dis[i]=-INF;
111         inq[i]=0;
112     }
113     int t=0,w=1;
114     q[1]=S; dis[S]=0; inq[S]=1;
115     while(t<w)
116     {
117         t++; int u=q[t%(s+5)]; inq[u]=0;
118         int e=head[u];
119         while(e)
120         {
121             int v=vet[e];
122             if(len1[e]&&dis[u]+len2[e]>dis[v])
123             {
124                 dis[v]=dis[u]+len2[e];
125                 pre[v][0]=u;
126                 pre[v][1]=e;
127                 if(!inq[v])
128                 {
129                     w++; q[w%(s+5)]=v; inq[v]=1;
130                 }
131             }
132             e=nxt[e];
133         }
134     }
135     if(dis[T]==-INF) return 0;
136     return 1;
137 }
138 
139 void mcf()
140 {
141     int k=T;
142     int t=INF;
143     while(k!=S)
144     {
145         intfor e = [k] [ 1 ];
146          t = min (T, len1 [e]);
147          a = the [k] [ 0 ];
148      }
 149      ANS1 + = t;
150      k = N;
151      while (a! = S)
 152      {
 153          int e = the [k] [ 1 ];
154          len1 [e] - = t;
155          len1 [e ^ 1 ] + = t;
156          ANS2 + = t * len2 [e];
157          a = the [k] [ 0 ];
158     }
159 }
160 
161 void solve1()
162 {
163     addedge();
164     ans1=ans2=0;
165     while(spfa1()) mcf();
166     printf("%d\n",ans2);
167 }
168 
169 void solve2()
170 {
171     addedge();
172     ans1=ans2=0;
173     while(spfa2()) mcf();
174     printf("%d\n",ans2);
175 }
176 
177 int main()
178 {
179     //freopen("1.in","r",stdin);
180     n=read(),m=read();
181     s=n+m,S=++s,T=++s;
182     rep(i,1,n) a[i]=read();
183     rep(i,1,m) b[i]=read();
184     rep(i,1,n)
185      rep(j,1,m) c[i][j]=read();
186     solve1();
187     solve2();
188 }