POJ 2516 Minimum Cost 最小费用流

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
",ans);
        else    printf("-1
");
    }
    return 0;
}