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=0; 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=0;i<NV;++i){
        vis[i]=0;
        d[i]=INF;
    }
    vis[vs]=1;
    d[vs]=0;
    queue<int> que;
    que.push(vs);
    while(!que.empty()){
        int u=que.front(); que.pop();
        for(int i=head[u]; i!=-1; 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]=1;
                    que.push(v);
                }
            }
        }
        vis[u]=0;
    }
    return d[vt]!=INF;
}
int MCMF(){
    int res=0;
    while(SPFA()){
        int flow=INF,cost=0;
        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]^1].cap+=flow;
            cost+=flow*edge[pre[u]].cost;
        }
        res+=cost;
    }
    return res;
}

int n,m,k,mat[55][55][55],to[55][55],from[55][55];
bool isOK(int *need,int *supply){
    for(int i=0; i<k; ++i) if(need[i]>supply[i]) return 0;
    return 1;
}
int main(){
    int a;
    while(~scanf("%d%d%d",&n,&m,&k) && (n||m||k)){
        int sum=0;
        int supply[55]={0},need[55]={0};
        for(int i=0; i<n; ++i){
            for(int j=0; j<k; ++j){
                scanf("%d",&to[i][j]);
                need[j]+=to[i][j];
            }
        }
        for(int i=0; i<m; ++i){
            for(int j=0; j<k; ++j){
                scanf("%d",&from[i][j]);
                supply[j]+=from[i][j];
            }
        }
        for(int z=0; z<k; ++z){
            for(int x=0; x<n; ++x){
                for(int y=0; y<m; ++y) scanf("%d",&mat[z][x][y]);
            }
        }
        if(!isOK(need,supply)){
            puts("-1");
            continue;
        }

        vs=n+m; vt=vs+1; NV=n+m+2;
        int res=0;
        for(int good=0; good<k; ++good){
            NE=0;
            memset(head,-1,sizeof(head));

            for(int i=0; i<m; ++i) addEdge(vs,i,from[i][good],0);
            for(int i=0; i<n; ++i) addEdge(i+m,vt,to[i][good],0);            
            for(int i=0; i<n; ++i){
                for(int j=0; j<m; ++j) addEdge(j,i+m,INF,mat[good][i][j]);
            }
            res+=MCMF();
        }
        printf("%d
",res);
    }
    return 0;
}