POJ 2516 Minimum Cost 最小费用最大流 费用流

题目连接：http://poj.org/problem?id=2516

题意跟题目输入这里写的很明白：http://blog.csdn.net/lyy289065406/article/details/6742534

无非就是k种物品，直接对每种物品进行一次最大流就OK了~

#include <iostream>
#include <queue>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 350;
const int inf = 10000000;
struct node
{
    int u,v,cap,flow,cost;
};

vector<struct node> edge;
vector<int> g[maxn];

void init(int n)
{
    int i;
    for(i = 0;i <= n;i++)
    g[i].clear();
    edge.clear();

    return ;
}
void addedge(int u,int v,int cap,int flow,int cost)
{
    edge.push_back((struct node){u,v,cap,flow,cost});
    edge.push_back((struct node){v,u,0,flow,-cost});
    int m;
    m = edge.size();
    g[u].push_back(m-2);
    g[v].push_back(m-1);
}
int shop[55][55];
int supply[55][55];
int vis[maxn],a[maxn],pre[maxn],dis[maxn];
int spfa(int s,int t,int n,int &flow,int & cost)
{
    int i;
    queue <int >q;
    for(i = 0;i <= n;i++)
    dis[i] = inf;
    dis[s] = 0;
    pre[s] = 0;memset(vis,0,sizeof(vis));
    vis[s] = 1;
    a[s] = inf;

    int u,v;
    q.push(s);

    while(!q.empty())
    {
        u = q.front();
        q.pop();
        vis[u] = 0;

        for(i = 0;i < g[u].size();i++)
        {
            struct node& e = edge[g[u][i]];
            v = e.v;
            if(e.cap > e.flow &&dis[v] > dis[u] + e.cost)
            {
                dis[v] = dis[u]+e.cost;
                a[v] = min(a[u],e.cap-e.flow);
                pre[v] = g[u][i];
                if(!vis[v])
                {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    if(dis[t] == inf)
    return 0;

    flow += a[t];
    cost += dis[t]*a[t];

    u = t;
    while( u != s)
    {
        edge[pre[u]].flow += a[t];
        edge[pre[u]^1].flow -= a[t];
        u = edge[pre[u]].u;
    }
    return 1;
}

int MCMF(int s,int t,int n)
{
    int flow = 0,cost = 0;
    while(spfa(s,t,n,flow,cost));
    return cost;
}
int main()
{
    int n,m,k;
    while (scanf("%d %d %d",&n,&m,&k)&&(n||m||k))
    {
        int i,j;
        for(i = 1;i <= n;i++)
        {
            for(j = 1;j <= k;j++)
            cin>>shop[i][j];
        }

        for(j = 1;j <= m;j++)
        for(i = 1;i <= k;i++)
        cin>>supply[j][i];

        int leap = 1;
        for(i = 1;i <= k;i++)
        {
            int need,support;
            need = support = 0;

            for(j = 1;j <= m;j++)
            support += supply[j][i];

            for(j = 1;j <= n;j++)
            need += shop[j][i];

            if(support < need)
            {
                leap = 0;
                break;
            }
        }
        int ii;
        int ans = 0;
        for(ii = 1;ii <= k;ii++)
        {
            init(n+m+2);
            for(i = 1;i <= m;i++)
            addedge(0,i,supply[i][ii],0,0);

            for(i = 1;i <= n;i++)
            addedge(i+m,n+m+1,shop[i][ii],0,0);

            for(i = 1;i <= n;i++)
            for(j = 1;j <= m;j++)
            {
                int cost;
                cin>>cost;
                addedge(j,i+m,inf,0,cost);
            }
            if(leap)
            ans += MCMF(0,n+m+1,n+m+2);
        }
        if(leap)
        cout<<ans<<endl;
        else
        puts("-1");

    }

    return 0;
}