POJ2516 Minimum Cost（最小费用最大流）

一开始我把每个店主都拆成k个点，然后建图。。然后TLE。。

看题解= =哦，愚钝了，k个商品是独立的，可以分别跑k次最小费用最大流，结果就是k次总和。。

 #include<cstdio>
 #include<cstring>
 #include<queue>
 #include<algorithm>
 using namespace std;
 #define INF (1<<30)
 #define MAXN 111
 #define MAXM 111*111*2
 struct Edge{
     int u,v,cap,cost,next;
 }edge[MAXM];
 int head[MAXN];
 int NV,NE,vs,vt;

 void addEdge(int u,int v,int cap,int cost){
     edge[NE].u=u; edge[NE].v=v; edge[NE].cap=cap; edge[NE].cost=cost;
     edge[NE].next=head[u]; head[u]=NE++;
     edge[NE].u=v; edge[NE].v=u; edge[NE].cap=; edge[NE].cost=-cost;
     edge[NE].next=head[v]; head[v]=NE++;
 }
 bool vis[MAXN];
 int d[MAXN],pre[MAXN];
 bool SPFA(){
     for(int i=;i<NV;++i){
         vis[i]=;
         d[i]=INF;
     }
     vis[vs]=;
     d[vs]=;
     queue<int> que;
     que.push(vs);
     while(!que.empty()){
         int u=que.front(); que.pop();
         for(int i=head[u]; i!=-; i=edge[i].next){
             int v=edge[i].v;
             if(edge[i].cap && d[v]>d[u]+edge[i].cost){
                 d[v]=d[u]+edge[i].cost;
                 pre[v]=i;
                 if(!vis[v]){
                     vis[v]=;
                     que.push(v);
                 }
             }
         }
         vis[u]=;
     }
     return d[vt]!=INF;
 }
 int MCMF(){
     int res=;
     while(SPFA()){
         int flow=INF,cost=;
         for(int u=vt; u!=vs; u=edge[pre[u]].u){
             flow=min(flow,edge[pre[u]].cap);
         }
         for(int u=vt; u!=vs; u=edge[pre[u]].u){
             edge[pre[u]].cap-=flow;
             edge[pre[u]^].cap+=flow;
             cost+=flow*edge[pre[u]].cost;
         }
         res+=cost;
     }
     return res;
 }

 int n,m,k,mat[][][],to[][],from[][];
 bool isOK(int *need,int *supply){
     for(int i=; i<k; ++i) if(need[i]>supply[i]) return ;
     return ;
 }
 int main(){
     int a;
     while(~scanf("%d%d%d",&n,&m,&k) && (n||m||k)){
         int sum=;
         int supply[]={},need[]={};
         for(int i=; i<n; ++i){
             for(int j=; j<k; ++j){
                 scanf("%d",&to[i][j]);
                 need[j]+=to[i][j];
             }
         }
         for(int i=; i<m; ++i){
             for(int j=; j<k; ++j){
                 scanf("%d",&from[i][j]);
                 supply[j]+=from[i][j];
             }
         }
         for(int z=; z<k; ++z){
             for(int x=; x<n; ++x){
                 for(int y=; y<m; ++y) scanf("%d",&mat[z][x][y]);
             }
         }
         if(!isOK(need,supply)){
             puts("-1");
             continue;
         }

         vs=n+m; vt=vs+; NV=n+m+;
         int res=;
         for(int good=; good<k; ++good){
             NE=;
             memset(head,-,sizeof(head));

             for(int i=; i<m; ++i) addEdge(vs,i,from[i][good],);
             for(int i=; i<n; ++i) addEdge(i+m,vt,to[i][good],);
             for(int i=; i<n; ++i){
                 for(int j=; j<m; ++j) addEdge(j,i+m,INF,mat[good][i][j]);
             }
             res+=MCMF();
         }
         printf("%d\n",res);
     }
     return ;
 }