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H - Minimum Cost

时间:2015-11-24 22:52:05      阅读:185      评论:0      收藏:0      [点我收藏+]

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#include<iostream>
#include<stdio.h>
#include<vector>
#include<string.h>
#include<queue>
using namespace std;

const int maxn=1024;
const int INF=0x3f3f3f3f;

struct Edge{
    int from,to,cap,flow,cost;
    Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w){}
};

struct MCMF{
    int n,m;
    vector<Edge>edges;
    vector<int>G[maxn];
    int inq[maxn];//是否在队列中
    int d[maxn];//Bellman-Ford
    int p[maxn];//上一条弧
    int a[maxn];//可改进量

    void init(int n){
        this->n=n;
        for(int i=0;i<n;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));
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BellmanFord(int s,int t,int &flow,long long &cost){
        for(int i=0;i<n;i++)d[i]=INF;
        memset(inq,0,sizeof(inq));
        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+=(long long)d[t]*(long long)a[t];
        for(int u=t;u!=s;u=edges[p[u]].from){
            edges[p[u]].flow+=a[t];
            edges[p[u]^1].flow-=a[t];
        }
        return true;
    }

    //需要保证初始网络中没有负权圈
    int MincostMaxflow(int s,int t,long long &cost){
        int flow=0;cost=0;
        while(BellmanFord(s,t,flow,cost));
        return flow;
    }
}MM;

int main(){
    int N,M,K;//汇点个数,源点个数,物品种类
    int need[64][64];//need[i][j]表示第i个店主需要第j种商品的数量
    int totalneed[64];//
    int storage[64][64];//storage[i][j]表示第i个仓库里存储的第j种商品的数量
    int totalstorage[64];//
    int cost[64][64];//cost[i][j]表示当前这个商品从第j个仓库运送到第i个店主的单位价格
    bool flag;
    int mincost;
    int maxflow;
    long long cost;

    while(~scanf("%d%d%d",&N,&M,&K)){
        if(N==0&&M==0&&K==0)break;
        memset(totalneed,0,sizeof(totalneed));
        memset(totalstorage,0,sizeof(totalstorage));
        flag=true;
        mincost=0;

        for(int i=0;i<N;++i){
            for(int j=0;j<K;++j){
                scanf("%d",&need[i][j]);
                totalneed[j]+=need[i][j];
            }
        }
        for(int i=0;i<M;++i){
            for(int j=0;j<K;++j){
                scanf("%d",&storage[i][j]);
                totalstorage[j]+=storage[i][j];
            }
        }
        for(int i=0;i<K;++i){
            if(totalneed[i]>totalstorage[i]){
                flag=false;
                break;
            }
        }

        for(int k=0;k<K;++k){
            for(int i=0;i<N;++i){
                for(int j=0;j<M;++j){
                    scanf("%d",&cost[i][j]);
                }
            }
            if(flag==false)continue;

            build(cost);//from here

            maxflow=MM.MincostMaxflow(s,t,cost);
            if(maxflow<totalneed[k])
                flag=false;
        }

        if(flag==false)printf("-1\n");
        else printf("%lld\n",mincost);
    }
    return 0;
}

 

H - Minimum Cost

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原文地址:http://www.cnblogs.com/bofengyu/p/4993054.html

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