洛谷P4015 运输问题（费用流）

传送门

源点向仓库连费用$0$，流量为储量的边，商店向汇点连费用$0$，流量为需求的边，然后仓库向商店连流量$inf$，费用对应的边，跑个费用流即可

 //minamoto
 #include<iostream>
 #include<cstdio>
 #include<queue>
 #include<cstring>
 #define inf 0x3f3f3f3f
 using namespace std;
 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
 char buf[<<],*p1=buf,*p2=buf;
 inline int read(){
     #define num ch-'0'
     char ch;bool flag=;int res;
     while(!isdigit(ch=getc()))
     (ch=='-')&&(flag=true);
     for(res=num;isdigit(ch=getc());res=res*+num);
     (flag)&&(res=-res);
     #undef num
     return res;
 }
 const int N=,M=;
 int ver[M],Next[M],head[N],edge[M],flow[M],tot=;
 int dis[N],disf[N],n,m,s,t,ans,Pre[N],last[N],vis[N],x;
 queue<int> q;
 inline void add(int u,int v,int e,int f){
     ver[++tot]=v,Next[tot]=head[u],head[u]=tot,edge[tot]=e,flow[tot]=f;
     ver[++tot]=u,Next[tot]=head[v],head[v]=tot,edge[tot]=-e,flow[tot]=;
 }
 bool spfa_min(){
     memset(dis,0x3f,sizeof(dis));
     memset(vis,,sizeof(vis));
     memset(disf,0x3f,sizeof(disf));
     q.push(s),dis[s]=,vis[s]=,Pre[t]=-;
     while(!q.empty()){
         int u=q.front();q.pop();vis[u]=;
         for(int i=head[u];i;i=Next[i]){
             int v=ver[i];
             if(flow[i]>&&dis[v]>dis[u]+edge[i]){
                 dis[v]=dis[u]+edge[i],Pre[v]=u;
                 last[v]=i,disf[v]=min(disf[u],flow[i]);
                 if(!vis[v]) vis[v]=,q.push(v);
             }
         }
     }
     return Pre[t]!=-;
 }
 int dinic_min(){
     int maxflow=,mincost=;
     while(spfa_min()){
         int u=t;
         maxflow+=disf[t],mincost+=disf[t]*dis[t];
         while(u!=s){
             flow[last[u]]-=disf[t];
             flow[last[u]^]+=disf[t];
             u=Pre[u];
         }
     }
     return mincost;
 }
 bool spfa_max(){
     memset(dis,0xef,sizeof(dis));
     memset(vis,,sizeof(vis));
     memset(disf,0x3f,sizeof(disf));
     q.push(s),dis[s]=,vis[s]=,Pre[t]=-;
     while(!q.empty()){
         int u=q.front();q.pop();vis[u]=;
         for(int i=head[u];i;i=Next[i]){
             int v=ver[i];
             if(flow[i]>&&dis[v]<dis[u]+edge[i]){
                 dis[v]=dis[u]+edge[i],Pre[v]=u;
                 last[v]=i,disf[v]=min(disf[u],flow[i]);
                 if(!vis[v]) vis[v]=,q.push(v);
             }
         }
     }
     return Pre[t]!=-;
 }
 int dinic_max(){
     int maxflow=,maxcost=;
     while(spfa_max()){
         int u=t;
         maxflow+=disf[t],maxcost+=disf[t]*dis[t];
         while(u!=s){
             flow[last[u]]-=disf[t];
             flow[last[u]^]+=disf[t];
             u=Pre[u];
         }
     }
     return maxcost;
 }
 int main(){
     m=read(),n=read();
     s=,t=n+m+;
     for(int i=;i<=m;++i){
         int x=read();
         add(s,i,,x);
     }
     for(int i=m+;i<=n+m;++i){
         int x=read();
         add(i,t,,x);
     }
     for(int i=;i<=m;++i)
     for(int j=;j<=n;++j){
         int x=read();
         add(i,j+m,x,inf);
     }
     printf("%d\n",dinic_min());
     for(int i=;i<=tot;i+=) flow[i]+=flow[i^],flow[i^]=;
     printf("%d\n",dinic_max());
     return ;
 }