图论--网络流--费用流--POJ 2156 Minimum Cost

Description

Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 to K). Once shopkeepers order goods, Dearboy should arrange which supply place provide how much amount of goods to shopkeepers to cut down the total cost of transport.

It's known that the cost to transport one unit goods for different kinds from different supply places to different shopkeepers may be different. Given each supply places' storage of K kinds of goods, N shopkeepers' order of K kinds of goods and the cost to transport goods for different kinds from different supply places to different shopkeepers, you should tell how to arrange the goods supply to minimize the total cost of transport.

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, K (0 < N, M, K < 50), which are described above. The next N lines give the shopkeepers' orders, with each line containing K integers (there integers are belong to [0, 3]), which represents the amount of goods each shopkeeper needs. The next M lines give the supply places' storage, with each line containing K integers (there integers are also belong to [0, 3]), which represents the amount of goods stored in that supply place.

Then come K integer matrices (each with the size N * M), the integer (this integer is belong to (0, 100)) at the i-th row, j-th column in the k-th matrix represents the cost to transport one unit of k-th goods from the j-th supply place to the i-th shopkeeper.

The input is terminated with three "0"s. This test case should not be processed.

Output

For each test case, if Dearboy can satisfy all the needs of all the shopkeepers, print in one line an integer, which is the minimum cost; otherwise just output "-1".

Sample Input

1 3 3   
1 1 1
0 1 1
1 2 2
1 0 1
1 2 3
1 1 1
2 1 1

1 1 1
3
2
20

0 0 0
Sample Output
4
-1

这个直接把每件物品单出来考虑完事，单独建图跑就行，然后就很简单了。就变成了普通的费用流问题，那么建图套模板即可！

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
#define INF 1e9
using namespace std;
const int maxn=200+10;
 
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[50+5][50+5];       //need[i][j]表i顾客对j商品的需求量
int have[50+5][50+5];       //have[i][j]表i仓库对j商品的提供量
int cost[50+5][50+5][50+5]; //cost[x][i][j] 表j仓库到i顾客对x商品的单位运费
 
int main()
{
    while(scanf("%d%d%d",&n,&m,&k)==3 && n)
    {
        int goods[maxn];//货物需求量,用来判断货物是否足够
        int enough=true;//初始货物充足
        memset(goods,0,sizeof(goods));
 
        for(int i=1;i<=n;++i)
        for(int j=1;j<=k;++j)
        {
            scanf("%d",&need[i][j]);
            goods[j]+= need[i][j];
        }
 
        for(int i=1;i<=m;++i)
        for(int j=1;j<=k;++j)
        {
            scanf("%d",&have[i][j]);
            goods[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(goods[i]>0)//货物不足,不用计算了
        {
            enough=false;
            break;
        }
        if(!enough)//初始货物不足
        {
            printf("-1\n");
            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,j+m,INF,cost[g][j][i]);
            }
            min_cost += MM.Min_cost();
        }
        printf("%d\n",min_cost);
    }
    return 0;
}

 

来源：CSDN

作者：张俊浩

链接：https://blog.csdn.net/weixin_43627118/article/details/102994972