POJ 2516 Minimum Cost

每个物品分开做最小费用最大流。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn=1000+10;
const int INF=0x7FFFFFFF;
struct Edge
{
    int from,to,cap,flow,cost;
};
int n,m,len,s,t;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];
int Flag,Ans,Xu;
int N,M,K;
int People[55][55];
int Supply[55][55];
int Cost[55][55][55];

void init()
{
    for(int i=0; i<maxn; i++) G[i].clear();
    edges.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
    });
    len=edges.size();
    G[from].push_back(len-2);
    G[to].push_back(len-1);
}

bool BellmanFord(int s,int t,int &flow,int &cost)
{

    for(int i=0; i<maxn; i++) d[i]=INF;

    memset(inq,0,sizeof(inq));
    memset(p,-1,sizeof(p));

    d[s]=0;
    inq[s]=1;
    p[s]=0;
    a[s]=INF;

    queue<int>Q;
    Q.push(s);
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        inq[u]=0;
        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]=1;
                }
            }
        }
    }
    if(d[t]==INF) return false;
    flow+=a[t];
    cost+=d[t]*a[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;
}

void Mincost (int s,int t)
{
    int flow=0,cost=0;
    while(BellmanFord(s,t,flow,cost));
    if(flow!=Xu) Flag=0;
    else Ans=Ans+cost;
}

int main()
{
    while(~scanf("%d%d%d",&N,&M,&K))
    {
        if(!N&&!M&&!K) break;
        Flag=1;
        Ans=0;
        for(int i=1; i<=N; i++)
            for(int j=1; j<=K; j++)
                scanf("%d",&People[i][j]);
        for(int i=1; i<=M; i++)
            for(int j=1; j<=K; j++)
                scanf("%d",&Supply[i][j]);
        for(int k=1; k<=K; k++)
            for(int i=1; i<=N; i++)
                for(int j=1; j<=M; j++)
                    scanf("%d",&Cost[k][i][j]);
        for(int k=1; k<=K; k++)
        {
            init();s=0;t=M+M+N+1;Xu=0;
            for(int i=1; i<=M; i++) Addedge(s,i,INF,0);
            for(int i=1; i<=M; i++) Addedge(i,i+M,Supply[i][k],0);
            for(int i=1; i<=N; i++)
                for(int j=1; j<=M; j++)
                    Addedge(j+M,M+M+i,INF,Cost[k][i][j]);
            for(int i=1; i<=N; i++) Addedge(i+M+M,t,People[i][k],0);
            for(int i=1; i<=N; i++) Xu=Xu+People[i][k];
            Mincost(s,t);
            if(!Flag) break;
        }
        if(Flag) printf("%d
",Ans);
        else printf("-1
");
    }
    return 0;
}
原文地址:https://www.cnblogs.com/zufezzt/p/4741198.html