运输问题

题目描述：

https://www.luogu.org/problemnew/show/P4015

题解：

建图跑最小费用流。

对于最大费用，我们可以重新建边，并把权值变成相反数。

也称 最大费用最大流。

代码：

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 105
#define ll long long
inline int rd()
{
    int f=1,c=0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=10*c+ch-'0';ch=getchar();}
    return f*c;
}
int m,n,S,T,hed1[2*N],cnt1=-1,hed2[2*N],cnt2=-1;
const ll inf = 0x3f3f3f3f3f3f3f3fll;
struct EG
{
    int to,nxt;
    ll vl,w;
}e1[N*N*2],e2[N*N*2];
void ae(int f,int t,ll v,ll w)
{
    e1[++cnt1].to = t;
    e1[cnt1].vl = v;
    e1[cnt1].w = w;
    e1[cnt1].nxt = hed1[f];
    hed1[f] = cnt1;
    e2[++cnt2].to = t;
    e2[cnt2].vl = v;
    e2[cnt2].w = -w;
    e2[cnt2].nxt = hed2[f];
    hed2[f] = cnt2;
}
int vis[N*2],pre[N*2],fa[N*2];
ll dis[N*2],f[N*2];
bool spfa1()
{
    queue<int>q;
    q.push(S);
    memset(dis,0x3f,sizeof(dis));
    memset(f,0,sizeof(f));
    dis[S]=0;f[S]=inf;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u]=0;
        for(int j=hed1[u];~j;j=e1[j].nxt)
        {
            int to = e1[j].to;
            if(e1[j].vl&&dis[to]>dis[u]+e1[j].w)
            {
                dis[to]=dis[u]+e1[j].w;
                f[to]=min(f[u],e1[j].vl);
                pre[to]=j;fa[to]=u;
                if(!vis[to])
                {
                    vis[to]=1;
                    q.push(to);
                }
            }
        }
    }
    return dis[T]!=inf;
}
ll mcmf1()
{
    ll ret = 0;
    while(spfa1())
    {
        ret+=dis[T]*f[T];
        int tmp = T;
        while(tmp!=S)
        {
            e1[pre[tmp]].vl-=f[T];
            e1[pre[tmp]^1].vl+=f[T];
            tmp=fa[tmp];
        }
    }
    return ret;
}
bool spfa2()
{
    queue<int>q;
    q.push(S);
    memset(dis,0x3f,sizeof(dis));
    memset(f,0,sizeof(f));
    dis[S]=0;f[S]=inf;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u]=0;
        for(int j=hed2[u];~j;j=e2[j].nxt)
        {
            int to = e2[j].to;
            if(e2[j].vl&&dis[to]>dis[u]+e2[j].w)
            {
                dis[to]=dis[u]+e2[j].w;
                f[to]=min(f[u],e2[j].vl);
                pre[to]=j;fa[to]=u;
                if(!vis[to])
                {
                    vis[to]=1;
                    q.push(to);
                }
            }
        }
    }
    return dis[T]!=inf;
}
ll mcmf2()
{
    ll ret = 0;
    while(spfa2())
    {
        ret+=dis[T]*f[T];
        int tmp = T;
        while(tmp!=S)
        {
            e2[pre[tmp]].vl-=f[T];
            e2[pre[tmp]^1].vl+=f[T];
            tmp=fa[tmp];
        }
    }
    return ret;
}

int main()
{
    memset(hed1,-1,sizeof(hed1));
    memset(hed2,-1,sizeof(hed2));
    m=rd(),n=rd();
    S=0,T=m+n+1;
    for(int a,i=1;i<=m;i++)
    {
        a=rd();
        ae(S,i,1ll*a,0);
        ae(i,S,0,0);
    }
    for(int b,i=1;i<=n;i++)
    {
        b=rd();
        ae(i+m,T,1ll*b,0);
        ae(T,i+m,0,0);
    }
    for(int i=1;i<=m;i++)
    {
        for(int c,j=1;j<=n;j++)
        {
            c=rd();
            ae(i,j+m,inf,c);
            ae(j+m,i,0,-c);
        }
    }
    ll ans1 = mcmf1();
    ll ans2 = -mcmf2();
    printf("%lld
%lld
",ans1,ans2);
    return 0;
}