POJ 2516 Minimum Cost 费用流

Posted 2021-01-01 zhenghanghu

建模比较难想。。

 

#include<iostream>
#include<deque>
#include<vector>
#include<cstring>
#include<cmath>
#define INF 2e9
using namespace std;

int T,ans;

struct edge{
    int v,cap,reverse,cost;
};
vector<int> edges[1005];//邻接表 
vector<edge> bian;//所有的边都存在里面

void addedge(int u,int v,int cost,int cap){//u到v一条边，再加一条反向边 
    edge e;
    e.cap=cap;
    e.v=v;
    e.cost=cost;
    e.reverse=bian.size()+1;//这条边的反边将建在这条边之后（这条边建完后在bian.size()的位置） 
    bian.push_back(e);
    edges[u].push_back(bian.size()-1);
    
    e.cap=0;
    e.v=u;
    e.cost=-cost;
    e.reverse=bian.size()-1;
    bian.push_back(e);
    edges[v].push_back(bian.size()-1);
}

int dist[1005],pre[1005];//pre[i]代表走的哪条边到的i点 

bool spfa(){
    for(int i=0;i<=T;i++) dist[i]=INF;
    for(int i=0;i<=T;i++) pre[i]=-1;
        
    deque<int> q;
    dist[0]=0; pre[0]=-1; q.push_back(0);
    while( !q.empty() ){
        int u = q.front(); q.pop_front();
        for(int i=0;i<edges[u].size();i++){//所有以u为起点的边的边的索引 
            edge &e = bian[ edges[u][i] ];
            int v=e.v;
            if( e.cap>0 && dist[u]+e.cost<dist[v] ){
                dist[v] = dist[u]+e.cost;
                pre[v]=edges[u][i];
                q.push_back(v);
            }
        }
    }
    
    if(dist[T]==INF) return false;
    return true;
}

int EK(){
    int max_flow=0;
    while( spfa() ){
        //cout<<"suc"<<endl;
        int u=T,minflow=INF;//在终点位置
        while( u!=0 ){
            edge &e = bian[ pre[u] ];
            minflow=min(minflow,e.cap);
            u = bian[ e.reverse ].v;
        }
        ans+=minflow*dist[T];
        max_flow+=minflow;
        u=T;
        while( u!=0 ){
            edge &e = bian[ pre[u] ];
            e.cap-=minflow;
            bian[ e.reverse ].cap+=minflow;
            u = bian[ e.reverse ].v;
        }
    }
    return max_flow;
}

int order[55][55],supply[55][55];//order是第i个商家对第j个物品的需求  supply第i个supply对第j个物品的库存 

int main(){
    int n,m,k;        
        
    while(1){
        cin>>n>>m>>k;
        if(n==0 && m==0 && k==0) break;
            
        ans=0;
        
        //supply：1-M  shopkeeper：M+1 - M+N
        T = n+m+1; 
        //k个物品间是相互独立的，所以等于跑k次费用流 
        for(int i=1;i<=n;i++)
            for(int j=1;j<=k;j++) cin>>order[i][j];
        
        for(int i=1;i<=m;i++)
            for(int j=1;j<=k;j++) cin>>supply[i][j];
        
        bool flag=true; 
        for(int k1=1;k1<=k;k1++){//k1个商品 
            bian.clear();
            for(int i=0;i<=T;i++) edges[i].clear();
            
            for(int i=1;i<=n;i++){
                for(int j=1;j<=m;j++){//第j个供应商给i商店供应k商品的代价 
                    int cost; cin>>cost;
                    addedge(j,m+i,cost,INF);
                }
            }
            //supply到shop解决了
            for(int i=1;i<=n;i++) addedge(m+i,T,0,order[i][k1]);//shop到汇点
            for(int i=1;i<=m;i++) addedge(0,i,0,supply[i][k1]);
            
            int sum=0;
            for(int i=1;i<=n;i++) sum+=order[i][k1];
            
            if( EK()!=sum ) flag=false;
        }
        if( flag ) cout<<ans<<endl;
        else cout<<"-1"<<endl;
    
    }
    
    return 0;    
}