每个物品分开做最小费用最大流。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std; const int maxn=+;
const int INF=0x7FFFFFFF;
struct Edge
{
int from,to,cap,flow,cost;
};
int n,m,len,s,t;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];
int Flag,Ans,Xu;
int N,M,K;
int People[][];
int Supply[][];
int Cost[][][]; void init()
{
for(int i=; i<maxn; i++) G[i].clear();
edges.clear();
} void Addedge(int from,int to,int cap,int cost)
{
edges.push_back((Edge)
{
from,to,cap,,cost
});
edges.push_back((Edge)
{
to,from,,,-cost
});
len=edges.size();
G[from].push_back(len-);
G[to].push_back(len-);
} bool BellmanFord(int s,int t,int &flow,int &cost)
{ for(int i=; i<maxn; i++) d[i]=INF; memset(inq,,sizeof(inq));
memset(p,-,sizeof(p)); d[s]=;
inq[s]=;
p[s]=;
a[s]=INF; queue<int>Q;
Q.push(s);
while(!Q.empty())
{
int u=Q.front();
Q.pop();
inq[u]=;
for(int i=; i<G[u].size(); i++)
{
Edge& e=edges[G[u][i]];
if(e.cap>e.flow&&d[e.to]>d[u]+e.cost)
{
d[e.to]=d[u]+e.cost;
p[e.to]=G[u][i];
a[e.to]=min(a[u],e.cap-e.flow);
if(!inq[e.to])
{
Q.push(e.to);
inq[e.to]=;
}
}
}
}
if(d[t]==INF) return false;
flow+=a[t];
cost+=d[t]*a[t];
int u=t;
while(u!=s)
{
edges[p[u]].flow+=a[t];
edges[p[u]^].flow-=a[t];
u=edges[p[u]].from;
}
return true;
} void Mincost (int s,int t)
{
int flow=,cost=;
while(BellmanFord(s,t,flow,cost));
if(flow!=Xu) Flag=;
else Ans=Ans+cost;
} int main()
{
while(~scanf("%d%d%d",&N,&M,&K))
{
if(!N&&!M&&!K) break;
Flag=;
Ans=;
for(int i=; i<=N; i++)
for(int j=; j<=K; j++)
scanf("%d",&People[i][j]);
for(int i=; i<=M; i++)
for(int j=; j<=K; j++)
scanf("%d",&Supply[i][j]);
for(int k=; k<=K; k++)
for(int i=; i<=N; i++)
for(int j=; j<=M; j++)
scanf("%d",&Cost[k][i][j]);
for(int k=; k<=K; k++)
{
init();s=;t=M+M+N+;Xu=;
for(int i=; i<=M; i++) Addedge(s,i,INF,);
for(int i=; i<=M; i++) Addedge(i,i+M,Supply[i][k],);
for(int i=; i<=N; i++)
for(int j=; j<=M; j++)
Addedge(j+M,M+M+i,INF,Cost[k][i][j]);
for(int i=; i<=N; i++) Addedge(i+M+M,t,People[i][k],);
for(int i=; i<=N; i++) Xu=Xu+People[i][k];
Mincost(s,t);
if(!Flag) break;
}
if(Flag) printf("%d\n",Ans);
else printf("-1\n");
}
return ;
}