题解 P4015 【运输问题】

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$Description$

$n$个仓库$m$个商店.$i$仓库有$a_i$单位货物，$i$商店需要$b_i$单位货物。$i$仓库到$j$商店的运费是$c_{i,j} /\text{单位}$。问最小运输费用和最大运输费用。

$Solution$

费用流水题

对于每个仓库$i$,每个商店$j$

$s \stackrel{a_{i}, 0}{\longrightarrow}i$

$i \stackrel{inf, c_{i,j}}{\longrightarrow}i$

$j \stackrel{b_{i}, 0}{\longrightarrow}t$

分别跑最小费用最大流和最大费用最大流即可

$Code$

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#include <bits/stdc++.h>
#define ll long long
#define re register
#define inf 0x3f3f3f3f
#define N 2000
using namespace std;
struct edge{
    int dis,w,to,next;
}e[1000067];
inline int read(){
    int x=0,w=0;char ch=getchar();
    while (!isdigit(ch))w|=ch=='-',ch=getchar();
    while (isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
    return w?-x:x;
}
int cnt=1,head[N],dis[N],vis[N],inque[N],cost,n,m,s,t,a[N],b[N],c[N][N];
inline void add(int u,int v,int d,int w){
    e[++cnt].to=v;
    e[cnt].next=head[u];
    e[cnt].dis=d;
    e[cnt].w=w;
    head[u]=cnt;
    e[++cnt].to=u;
    e[cnt].next=head[v];
    e[cnt].dis=0;
    e[cnt].w=-w;
    head[v]=cnt;
}
bool spfa1(){
    memset(dis,0x3f,sizeof(dis));
    queue<int>q;q.push(s);
    dis[s]=0;
    while (!q.empty()){
        int u=q.front();q.pop();inque[u]=0;
        for (int i=head[u];i;i=e[i].next){
            int v=e[i].to;
            if (e[i].dis&&dis[v]>dis[u]+e[i].w){
                dis[v]=dis[u]+e[i].w;
                if (!inque[v])
                    q.push(v),inque[v]=1;
            }
        }
    }
    return dis[t]!=inf;
}
bool spfa2(){
    for (int i=s;i<=t;++i)dis[i]=-inf;
    queue<int>q;q.push(s);
    dis[s]=0;
    while (!q.empty()){
        int u=q.front();q.pop();inque[u]=0;
        for (int i=head[u];i;i=e[i].next){
            int v=e[i].to;
            if (e[i].dis&&dis[v]<dis[u]+e[i].w){
                dis[v]=dis[u]+e[i].w;
                if (!inque[v])
                    q.push(v),inque[v]=1;
            }
        }
    }
    return dis[t]!=-inf;
}
int dfs(int u,int mn){
    vis[u]=1;
    if (u==t)return mn;
    int used=0,mi;
    for (int i=head[u];i;i=e[i].next){
        int v=e[i].to;
        if ((!vis[v]||v==t)&&e[i].dis&&dis[v]==dis[u]+e[i].w)
            if (mi=dfs(v,min(e[i].dis,mn-used))){
                e[i].dis-=mi;
                e[i^1].dis+=mi;
                used+=mi;
                cost+=mi*e[i].w;
                if (used==mn)break;
            }
    }
    return used;
}
void Dinic1(){
    while (spfa1()){
        vis[t]=1;
        while (vis[t]){
            memset(vis,0,sizeof(vis));
            dfs(s,inf);
        }
    }
}
void Dinic2(){
    while (spfa2()){
        vis[t]=1;
        while (vis[t]){
            memset(vis,0,sizeof(vis));
            dfs(s,inf);
        }
    }
}
signed main(){
    n=read(),m=read(),s=0,t=n+m+1;
    for (int i=1;i<=n;++i)a[i]=read(),add(s,i,a[i],0);
    for (int i=1;i<=m;++i)b[i]=read(),add(i+n,t,b[i],0);
    for (int i=1;i<=n;++i)
        for (int j=1;j<=m;++j)
            c[i][j]=read(),add(i,j+n,inf,c[i][j]);
    Dinic1();
    printf("%d\n",cost);
    memset(head,0,sizeof(head));cnt=1;cost=0;
    for (int i=1;i<=n;++i)add(s,i,a[i],0);
    for (int i=1;i<=m;++i)add(i+n,t,b[i],0);
    for (int i=1;i<=n;++i)
        for (int j=1;j<=m;++j)
            add(i,j+n,inf,c[i][j]);
    Dinic2();
    printf("%d\n",cost);
    return 0;
}