POJ2516 Minimum Cost Minimum Cost Maximum []

Topic links: http://poj.org/problem?id=2516

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions:19928		Accepted: 7064

Description

Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 to K). Once shopkeepers order goods, Dearboy should arrange which supply place provide how much amount of goods to shopkeepers to cut down the total cost of transport. 

It's known that the cost to transport one unit goods for different kinds from different supply places to different shopkeepers may be different. Given each supply places' storage of K kinds of goods, N shopkeepers' order of K kinds of goods and the cost to transport goods for different kinds from different supply places to different shopkeepers, you should tell how to arrange the goods supply to minimize the total cost of transport.

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, K (0 < N, M, K < 50), which are described above. The next N lines give the shopkeepers' orders, with each line containing K integers (there integers are belong to [0, 3]), which represents the amount of goods each shopkeeper needs. The next M lines give the supply places' storage, with each line containing K integers (there integers are also belong to [0, 3]), which represents the amount of goods stored in that supply place. 

Then come K integer matrices (each with the size N * M), the integer (this integer is belong to (0, 100)) at the i-th row, j-th column in the k-th matrix represents the cost to transport one unit of k-th goods from the j-th supply place to the i-th shopkeeper. 

The input is terminated with three "0"s. This test case should not be processed.

Subject to the effect: n businessman, m warehouse, each warehouse, there are k kinds of goods. Given the needs of each merchant for goods, given the amount of each warehouse to store goods, giving each warehouse to each merchant each unit of goods spending. Seek minimum cost while meeting the needs of businessmen.

Ideas:

1. This question is entered it is made uncomfortable, although the last was written out. I do not want to explain to explain the code inside.

2. The key is that each cargo is independent, there is no association, it is possible to carry out separately for each item MCMF (). Final results will be superimposed.

3. The composition is easy to think, for the current cargo warehouse connected to each source point edge with a capacity of warehouse storage capacity of the cargo, the cost is zero. Warehouse to each merchant even side, capacity inf, to spend for the kinds of goods spent the businessman units. Each businessmen even to edge meeting point for businessmen capacity demand, the cost is zero.

code show as below:

  . 1 #include <stdio.h>
   2 #include < String .h>
   . 3 #include <Queue>
   . 4 #include <algorithm>
   . 5  #define MEM (A, B) Memset (A, B, the sizeof (A))
   . 6  const  int MAXN = 300 ;
   . 7  const  int INF = 0x3f3f3f3f ;
   . 8  the using  namespace STD;
   . 9  
10  int n-, m, k; // n-m merchant warehouses with k respectively each warehouse goods 
. 11  int NUM [MAXN] [MAXN ]; // represents the i j th individual needs of goods number 
12  intc_num [MAXN] [MAXN]; // represents the number of i-th j th warehouse storing goods 
13 is  int VIS [ 2 * MAXN], DIS [ 2 * MAXN], Flow [ 2 * MAXN];
 14  int pre [ 2 * MAXN], Last [ 2 * MAXN], minCost, MaxFlow;
 15 Queue < int > Q;
 16  
. 17  struct Edge
 18 is  {
 . 19      int to, Next, Flow, DIS;
 20 is } Edge [ 2 * MAXN + MAXN * MAXN] ;
 21 is  int head [MAXN], CNT;
 22 is 
 23 void add(int a, int b, int c, int d)
 24 {
 25     cnt ++;
 26     edge[cnt].to = b;
 27     edge[cnt].flow = c;
 28     edge[cnt].dis = d;
 29     edge[cnt].next = head[a];
 30     head[a] = cnt;
 31 }
 32 
 33 bool spfa(int st, int ed)
 34 {
 35     mem(dis, inf), mem(flow, inf), mem(vis, 0);
 36     Q.push(st);
 37     vis[st] = 1;
 38     dis[st] = 0;
 39     pre[ed] = -1;
 40     while(!Q.empty())
 41     {
 42         int now = Q.front();
 43         Q.pop();
 44         vis[now] = 0;
 45         for(int i = head[now]; i != -1; i = edge[i].next)
 46         {
 47             int to = edge[i].to;
 48             if(edge[i].flow > 0 && dis[to] > dis[now] + edge[i].dis)
 49             {
 50                 dis[to] = dis[now] + edge[i].dis;
 51                 pre[to] = now;
 52                 last[to] = i;
 53                 flow[to] = min(flow[now], edge[i].flow);
 54                 if(!vis[to])
 55                 {
 56                     vis[to] = 1;
 57                     Q.push(to);
 58                 }
 59             }
 60         }
 61     }
 62     return pre[ed] != -1;
 63 }
 64 
 65 void MCMF()
 66 {
 67     int st = 0, ed = m + n + 1;
 68     maxflow = 0, mincost = 0;
 69     while(spfa(st, ed))
 70     {
 71         int now = ed;
 72         maxflow += flow[ed];
 73         mincost += flow[ed] * dis[ed];
 74         while(now != st)
 75         {
 76             edge[last[now]].flow -= flow[ed];
 77             edge[last[now] ^ 1].flow += flow[ed];
 78             now = pre[now];
 79         }
 80     }
 81 }
 82 
 83 int main()
 84 {
 85     while(scanf("%d%d%d"!, & N-, & m, & K) = the EOF)
 86      {
 87          int In Flag = . 1 ; // number of items stored in the warehouse is sufficient 
88          IF (n-== 0 && m == 0 && K == 0 )
 89              BREAK ;
 90          for ( int I = . 1 ; I <= n-; I ++ ) 
 91 is              for ( int J = . 1 ; J <= K; J ++ )
 92                  Scanf ( " % D " , & NUM [I] [J ]);
 93          for( Int I = . 1 ; I <= m; I ++ )
 94              for ( int J = . 1 ; J <= K; J ++ )
 95                  Scanf ( " % D " , & [I] c_num [J]);
 96          for ( int I = . 1 ; I <= K; I ++) // Laid enumerate each determination goods 
97          {
 98              int X = 0 , Y = 0 ;
 99              for ( int J = . 1 ; J <= n- ; J ++) // enum people 
100                 + = X NUM [J] [I];
 101              for ( int J = . 1 ; J <= m; J ++) // Enumeration warehouse 
102                  Y + = c_num [J] [I];
 103              IF (X> Y)
 104                  in Flag = 0 ;
 105          }
 106          int ANS = 0 ;
 107          for ( int K = . 1 ; K <= K; K ++) // K matrix where each enumeration cost item 
108          {
 109              CNT = - . 1 , MEM (head, -. 1 );
 110              for ( int I = . 1 ; I <= m; I ++) // source built into the sides of the warehouse 
111              {
 112                  the Add ( 0 , I, c_num [I] [K], 0 ); // super source 0 
113                  the Add (I, 0 , 0 , 0 );
 114              }
 115              for ( int I = . 1 ; I <= n-; I ++) // warehouse built to represent the human side of the warehouse at which article the unit cost person 
1 16              {
 117                  for ( int j = 1; j <= m; j ++)
118                 {
119                     int x;
120                     scanf("%d", &x);
121                     if(flag == 0)
122                         continue;
123                     add(j, m + i, inf, x);
124                     add(m + i, j, 0, -x);
125                 }
126             }
127             for(int i = 1; i <= n; i ++)//People to super sink m + n + 1 building side 
128              {
 129                  the Add (m + I, m + n-+ . 1 , NUM [I] [K], 0 );
 130.                  the Add (m + n-+ . 1 , m + I , 0 , 0 );
 131 is              }
 132              MCMF ();
 133              ANS + = minCost;
 134          }
 135          IF (In Flag)
 136              the printf ( " % D \ n- " , ANS);
 137          the else 
138              the printf ( " -1 \ n- ");
139     }
140     return 0;
141 }

POJ2516