poj 2516 Minimum Cost(费用流)

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 17714		Accepted: 6225

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.

Output

For each test case, if Dearboy can satisfy all the needs of all the shopkeepers, print in one line an integer, which is the minimum cost; otherwise just output "-1".

Sample Input

1 3 3
1 1 1
0 1 1
1 2 2
1 0 1
1 2 3
1 1 1
2 1 1

1 1 1
3
2
20

0 0 0

Sample Output

4
-1

题意：给出n个店，m个供应商，k种货，然后是n行，每行是店对每种货物的需求量，接着有k个n*m的矩阵，表示j对i运一件货的花费是多少，求所有店都可以满足需求的最小花费 ，如果存在店不满足需求，输出-1

费用流问题，将原图拆成k个图，每张图如下所示（只写了流量）

花费的话只有供应商和店之间有花费



可以在之前提前判断一下每个货物的总量是否满足，优化一些时间

#include <cstdio>
#include <cstring>
#include <cctype>
#include <stdlib.h>
#include <string>
#include <map>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
const int N=5000+20;
const int M=50000+20;

int top;//当前边下标
int dis
,pre
;//源点到点i的最小距离，pre[i]记录前驱
bool vis
;//标记数组
int maxflow;
struct vertex
{
int first;
} V
;
struct Edge
dc80

{
int v,next;
int cap,flow,cost;
} E[M*2];

void init()
{
mem(V,-1);
top=0;
maxflow=0;
}

void add_edge(int u,int v,int c,int cost)
{
E[top].v=v;
E[top].cap=c;
E[top].flow=0;
E[top].cost=cost;
E[top].next=V[u].first;
V[u].first=top++;
}
void add(int u,int v,int c,int cost)
{
add_edge(u,v,c,cost);
add_edge(v,u,0,-cost);
}
bool spfa(int s,int t)
{
int i,u,v;
queue<int>q;
mem(vis,false);
mem(pre,-1);
mem(dis,inf);
vis[s]=true;
dis[s]=0;
q.push(s);
while(!q.empty())
{
u=q.front();
q.pop();
vis[u]=false;
for(int i=V[u].first; i!=-1; i=E[i].next)
{
v=E[i].v;
if(E[i].cap>E[i].flow&&dis[v]>dis[u]+E[i].cost)
{
dis[v]=dis[u]+E[i].cost;
pre[v]=i;
if(!vis[v])
{
q.push(v);
vis[v]=true;
}
}
}
}
if(dis[t]==inf)
return false;
return true;
}
int MCMF(int s,int t)//minCostMaxFlow
{
int d;
int i,mincost=0;//maxflow当前最大流量，mincost当前最小费用
while(spfa(s,t))//表示找到了从s到t的最小费用路
{
d=inf;
for(int i=pre[t]; i!=-1; i=pre[E[i^1].v]) //遍历反向边
d=min(d,E[i].cap-E[i].flow);
maxflow+=d;//更新最大流
//printf("%d\n",d);
for(int i=pre[t]; i!=-1; i=pre[E[i^1].v]) //增广路上正向边流量+d,反向边流量-d
{
E[i].flow+=d;
E[i^1].flow-=d;
}
mincost+=dis[t]*d;//dis[t]为该路径上单位流量费用之和
}
return mincost;
}
int e1[55][55],e2[55][55],w[55][55],check[55];
int main()
{

int n,m,k,i,j,o;
while(scanf("%d%d%d",&n,&m,&k)&&n+m+k)
{
int ans=0;
mem(check,0);
for(i=1; i<=n; i++)
for(j=1; j<=k; j++)
{
scanf("%d",&e1[i][j]);
check[j]+=e1[i][j];
}
for(i=1; i<=m; i++)
for(j=1; j<=k; j++)
{
scanf("%d",&e2[i][j]);
}

bool flag=0;
for(o=1; o<=k; o++)
{
//源-》供货
init();
for(i=1; i<=n; i++)
{
//供货-》店
for(j=1; j<=m; j++)
{
scanf("%d",&w[j][i]);
add(j,m+i,e2[j][o],w[j][i]);
}
}
if(flag==1)continue;
for(i=1;i<=m;i++)
add(0,i,e2[i][o],0);
for(i=1;i<=n;i++)
add(m+i,n+m+1,e1[i][o],0);
int now=MCMF(0,m+n+1);
if(maxflow!=check[o])
{
flag=1;
continue;
}
else
ans+=now;

}
if(flag)
puts("-1");
else
printf("%d\n",ans);
}
return 0;
}