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POJ 2516 Minimum Cost 最小费用流

2015-08-26 20:58 169 查看
这题意恶心得我百度了。看小悠写的看懂的*_*

各个商品的运输的代价计算是独立的,所以可以将他们分别处理

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int INF=0x7f7f7f7f;
const int maxn=100008;
const int maxm=58;
struct fuck{
int u,v,cap,next,cost;
}edge[maxn];
int head[maxm<<4];
int tol;
int mpn[maxm][maxm],mpm[maxm][maxm],mpcost[maxm][maxm][maxm];
void init()
{
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int c)
{
edge[tol].u=u;
edge[tol].v=v;
edge[tol].cap=w;
edge[tol].next=head[u];
edge[tol].cost=c;
head[u]=tol++;
edge[tol].u=v;
edge[tol].v=u;
edge[tol].cap=0;
edge[tol].next=head[v];
edge[tol].cost=-c;
head[v]=tol++;
}
int micost,max_flow,last;
int dis[maxm<<4],pre[maxn<<4];
bool vis[maxm<<4];
bool spfa(int sour,int sink)
{
queue<int>    q;
q.push(sour);
memset(vis,false,sizeof(vis));
memset(dis,INF,sizeof(dis));
dis[sour]=0;vis[sour]=true;pre[sour]=0;
int u,v,i;
while(!q.empty())
{
u=q.front();q.pop();
vis[u]=false;
for(i=head[u];i!=-1;i=edge[i].next)
{
v=edge[i].v;
if(edge[i].cap>0&&dis[v]>dis[u]+edge[i].cost)
{
dis[v]=dis[u]+edge[i].cost;
pre[v]=i;
if(!vis[v])
{
q.push(v);
vis[v]=true;
}
}
}
}
if(dis[sink]>=INF)    return false;
return true;
}
void mincost_flow(int sour,int sink)
{
max_flow=0;micost=0;
int i;
while(spfa(sour,sink))
{
int fl=INF;
for(i=sink;i!=sour;i=edge[pre[i]].u)
if(fl>edge[pre[i]].cap)
fl=edge[pre[i]].cap;
for(i=sink;i!=sour;i=edge[pre[i]].u)
{
edge[pre[i]].cap-=fl;
edge[pre[i]^1].cap+=fl;
}
max_flow+=fl;
micost+=dis[sink]*fl;
}
}
int main()
{
int i,j,k,jj,n,m,u,v,w;
while(scanf("%d%d%d",&n,&m,&k)==3)
{
if(n==0&&m==0&&k==0)    break;
bool flag=true;
int ans=0;
for(i=1;i<=n;i++)
for(j=1;j<=k;j++)
scanf("%d",&mpn[i][j]);
for(i=1;i<=m;i++)
for(j=1;j<=k;j++)
scanf("%d",&mpm[i][j]);
for(jj=1;jj<=k;jj++)
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
scanf("%d",&mpcost[jj][i][j]);
for(i=1;i<=k;i++)
{
init();
int sum=0;
for(j=1;j<=n;j++)//lingshoushang
if(mpn[j][i]>0)
{
addedge(j,n+m+1,mpn[j][i],0);
sum+=mpn[j][i];
}
for(j=1;j<=m;j++)//shangjia
if(mpm[j][i]>0)
addedge(0,j+n,mpm[j][i],0);
int sour=0,sink=n+m+1;
for(j=1;j<=n;j++)
{
if(mpn[j][i]>0)
for(jj=1;jj<=m;jj++)
if(mpm[jj][i]>0)
addedge(jj+n,j,mpm[jj][i],mpcost[i][j][jj]);
}
last=sour+1;
mincost_flow(sour,sink);
if(max_flow<sum)    flag=false;
ans+=micost;
}
if(flag)    printf("%d\n",ans);
else    printf("-1\n");
}
return 0;
}
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