POJ 2516 Minimum Cost(最小费用最大流)

大佬的博客：http://blog.youkuaiyun.com/u013480600/article/details/39026241

题意:

       给出n个客户对k种商品的需求量，又给出m个仓库对k种物品的存货量以及对k种物品从i仓库到j客户的一个物品的运费价格，让判断是否可以满足客户需求，然后就是如果满足求出最小的运费.

分析:

       首先我们必须判断m个仓库是否有足够的k种商品给n个客户,如果不足,那么明显就是不行的. 下面假设仓库的商品足够的话:

       对于每一种商品我们都算出满足满足顾客需求量的最小运费即可.所以我们对K种商品分开处理如下,假设当前处理第x种商品,建图如下:

       源点s编号0, m个仓库编号1到m, n个顾客编号m+1到m+n, 汇点编号m+n+1.

       从源点s到每个仓库i有边(s, i, 仓库i对商品x的存货量, 0)

       从每个仓库i到顾客j有边(i, j, INF, 仓库i到顾客j的单位X商品的运费)

       从每个顾客j到汇点t有边(j, t, 顾客j对X商品的需求量, 0)

       然后我们求最小费用最大流即可求出N个顾客对第X种商品的最小运费.

#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1e9
using namespace std;
const int maxn = 105;
struct Edge
{
    int from,to,cap,flow,cost;
    Edge(){}
    Edge(int f,int t,int c,int fl,int co):from(f),to(t),cap(c),flow(fl),cost(co){}
};
struct MCMF
{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inq[maxn];
    int d[maxn];
    int p[maxn];
    int a[maxn];

    void init(int n,int s,int t)
    {
        this->n=n, this->s=s, this->t=t;
        edges.clear();
        for(int i=0;i<n;++i) G[i].clear();
    }

    void AddEdge(int from,int to,int cap,int cost)
    {
        edges.push_back(Edge(from,to,cap,0,cost));
        edges.push_back(Edge(to,from,0,0,-cost));
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BellmanFord(int &flow, int &cost)
    {
        for(int i=0;i<n;++i) d[i]=INF;
        memset(inq,0,sizeof(inq));
        d[s]=0, a[s]=INF, inq[s]=true, p[s]=0;
        queue<int> Q;
        Q.push(s);
        while(!Q.empty())
        {
            int u=Q.front(); Q.pop();
            inq[u]=false;
            for(int i=0;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]=true; }
                }
            }
        }
        if(d[t]==INF) return false;
        flow +=a[t];
        cost +=a[t]*d[t];
        int u=t;
        while(u!=s)
        {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -=a[t];
            u = edges[p[u]].from;
        }
        return true;
    }
    int Min_cost()
    {
        int flow=0,cost=0;
        while(BellmanFord(flow,cost));
        return cost;
    }
}MM;
int n, m, k;
int need[55][55];
int have[55][55];
int cost[55][55][55];
int main()
{
    while(~scanf("%d%d%d", &n, &m, &k) && n)
    {
        int good[maxn];
        bool enough = true;
        memset(good, 0, sizeof(good));
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= k; j++)
            {
                scanf("%d", &need[i][j]);
                good[j] += need[i][j];
            }
        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= k; j++)
            {
                scanf("%d", &have[i][j]);
                good[j] -= have[i][j];
            }
        for(int h = 1; h <= k; h++)
            for(int i = 1; i <= n; i++)
                for(int j = 1; j <= m; j++)
                    scanf("%d", &cost[h][i][j]);
        for(int i = 1; i <= k; i++)
            if(good[i] > 0)
            {
                enough = false;
                break;
            }
        if(!enough)
        {
            puts("-1");
            continue;
        }
        int min_cost = 0;
        for(int g = 1; g <= k; g++)
        {
            int src = 0, dst = n+m+1;
            MM.init(n+m+2, src, dst);
            for(int i = 1; i <= m; i++)
                MM.AddEdge(src, i, have[i][g], 0);
            for(int i = 1; i <= n; i++)
                MM.AddEdge(m+i, dst, need[i][g], 0);
            for(int i = 1; i <= m; i++)
                for(int j = 1; j <= n; j++)
                    MM.AddEdge(i, m+j, INF, cost[g][j][i]);
            min_cost += MM.Min_cost();
        }
        printf("%d\n",min_cost);
    }
    return 0;
}