• 【网络流24题】运输问题(费用流)


    【网络流24题】运输问题(费用流)

    题面

    Cogs

    题解

    大水题。。。
    源点向仓库连,容量为货物量,费用为0
    仓库向商店连,容量INF,费用题目给出来了
    商店向汇点连,容量为需求量,费用为0
    简直裸的费用流

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #include<set>
    #include<map>
    #include<vector>
    #include<queue>
    using namespace std;
    #define MAX 5000
    #define MAXL 500000
    #define INF 1000000000
    inline int read()
    {
    	int x=0,t=1;char ch=getchar();
    	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    	if(ch=='-')t=-1,ch=getchar();
    	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    	return x*t;
    }
    struct Line
    {
        int v,next,w,fy;
    }e[MAXL];
    bool vis[MAX];
    int h[MAX],cnt=2;
    int a[MAX],b[MAX],c[MAX][MAX];
    inline void Add(int u,int v,int w,int fy)
    {
        e[cnt]=(Line){v,h[u],w,fy};h[u]=cnt++;
        e[cnt]=(Line){u,h[v],0,-fy};h[v]=cnt++;
    }
    int pe[MAX],pr[MAX],dis[MAX];
    int S,T,Cost,n,m,Flow,opt=1;
    bool SPFA()
    {
        memset(dis,63,sizeof(dis));
        queue<int> Q;
        Q.push(S);dis[S]=0;
        while(!Q.empty())
        {
            int u=Q.front();Q.pop();
            for(int i=h[u];i;i=e[i].next)
            {
                int v=e[i].v;
                if(e[i].w&&dis[v]>dis[u]+e[i].fy)
                {
                    dis[v]=dis[u]+e[i].fy;
                    pe[v]=i;pr[v]=u;
                    if(!vis[v])vis[v]=true,Q.push(v);
                }
            }
            vis[u]=false;
        }
        if(dis[T]>=INF)return false;
        int flow=INF;
        for(int i=T;i!=S;i=pr[i])flow=min(flow,e[pe[i]].w);
        for(int i=T;i!=S;i=pr[i])e[pe[i]].w-=flow,e[pe[i]^1].w+=flow;
        Cost+=opt*flow*dis[T];
        Flow+=flow;
        return true;
    }
    int main()
    {
    	freopen("tran.in","r",stdin);
    	freopen("tran.out","w",stdout);
    	m=read();n=read();
    	for(int i=1;i<=m;++i)a[i]=read();
    	for(int j=1;j<=n;++j)b[j]=read();
    	for(int i=1;i<=m;++i)
    		for(int j=1;j<=n;++j)
    			c[i][j]=read();
    	S=0;T=n+m+1;
    	
    	memset(h,0,sizeof(h));cnt=2;
    	for(int i=1;i<=m;++i)Add(S,i,a[i],0);
    	for(int j=1;j<=n;++j)Add(j+m,T,b[j],0);
    	for(int i=1;i<=m;++i)
    		for(int j=1;j<=n;++j)
    			Add(i,j+m,INF,c[i][j]);
    	Cost=Flow=0;opt=1;
    	while(SPFA());
    	printf("%d
    ",Cost);
    
    	memset(h,0,sizeof(h));cnt=2;
    	for(int i=1;i<=m;++i)Add(S,i,a[i],0);
    	for(int j=1;j<=n;++j)Add(j+m,T,b[j],0);
    	for(int i=1;i<=m;++i)
    		for(int j=1;j<=n;++j)
    			Add(i,j+m,INF,-c[i][j]);
    	Cost=Flow=0;opt=-1;
    	while(SPFA());
    	printf("%d
    ",Cost);
    
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cjyyb/p/8193411.html
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