NOI2012 美食节

http://www.lydsy.com/JudgeOnline/problem.php?id=2879

费用流。

我们发现，每个厨师做的倒数第k道菜对总等待时间的贡献为k*做这道菜的时间。

将每个厨师拆成P个点，第i个第表示这个厨师做倒数第i道菜。

设Vi,j表示第i个厨师做第j道菜的点。

Ui表示第i道菜。

构图：

S->Vi,j一条流量为1，费用为0的边；

Vi,j->Uk一条流量为1，费用为j*t[k][i]的边；

Ui->T一条流量为p[i]，费用为0的边。

但是数据范围比较大，我们可以动态加边。

友情题：SDOI2007修车

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj

using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define re(i,a,b)  for(i=a;i<=b;i++)
#define red(i,a,b) for(i=a;i>=b;i--)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

const DB EPS=1e-;
inline int sgn(DB x){if(abs(x)<EPS)return ;return(x>)?:-;}
const DB Pi=acos(-1.0);

inline int gint()
  {
        int res=;bool neg=;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return ;
        if(z=='-'){neg=;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());
        return (neg)?-res:res;
    }
inline LL gll()
  {
      LL res=;bool neg=;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return ;
        if(z=='-'){neg=;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());
        return (neg)?-res:res;
    }

const int maxN=;
const int maxM=;
const int maxP=;
const int INF=<<;

int N,M,P;
int p[maxN+];
int t[maxN+][maxM+];
int S,T,now,first[maxN+maxM*maxP+];
struct Tedge{int u,v,flow,cost,next;}edge[*(maxN+maxN*maxM*maxP+maxP*maxM)+];

inline void addedge(int u,int v,int flow,int cost)
  {
      now++;
      edge[now].u=u;
      edge[now].v=v;
      edge[now].flow=flow;
      edge[now].cost=cost;
      edge[now].next=first[u];
      first[u]=now;
  }
inline void insert(int u,int v,int flow,int cost){addedge(u,v,flow,cost);addedge(v,u,,-cost);}

int head,tail,que[maxN+maxM*maxP+];
int vis[maxN+maxM*maxP+],dis[maxN+maxM*maxP+],fromedge[maxN+maxM*maxP+];
inline int SPFA()
  {
      int i;
      re(i,,N+P*M+)vis[i]=,dis[i]=INF,fromedge[i]=-;
      dis[que[]=S]=;
      head=;tail=;
      vis[S]=;
      while(head!=tail)
        {
            int u=que[head++],v,flow,cost;if(head==T)head=;
            vis[u]=;
            for(i=first[u],v=edge[i].v,flow=edge[i].flow,cost=edge[i].cost;i!=-;i=edge[i].next,v=edge[i].v,flow=edge[i].flow,cost=edge[i].cost)
              if(flow> && dis[u]+cost<dis[v])
                {
                    dis[v]=dis[u]+cost;
                    fromedge[v]=i;
                    if(!vis[v])
                              {
                                  vis[que[tail]=v]=;
                                  if(dis[que[tail]]<dis[que[head]])swap(que[head],que[tail]);
                                  tail++;if(tail==T)tail=;
                                }
                }
        }
      return dis[T]!=INF;
  }
inline void work(int &res)
  {
      int i,x=INF,y,a,b;
      for(i=fromedge[T];i!=-;i=fromedge[edge[i].u])
          {
              upmin(x,edge[i].flow);
              if(edge[i].u==S)y=edge[i].v,a=(y-)/P+,b=y%P+;
            }
      for(i=fromedge[T];i!=-;i=fromedge[edge[i].u])edge[i].flow-=x,edge[i^].flow+=x,res+=x*edge[i].cost;
      re(i,,N)insert((a-)*P+b,P*M+i,,b*t[i][a]);
  }
inline int mincostmaxflow()
  {
      int res=;
        while(SPFA())work(res);
      return res;
  }

int main()
  {
      freopen("delicacy.in","r",stdin);
      freopen("delicacy.out","w",stdout);
      int i,j;
      N=gint();M=gint();
      mmst(first,-);now=-;
      re(i,,N)p[i]=gint(),P+=p[i];
        re(i,,N)re(j,,M)t[i][j]=gint();
        S=;T=P*M+N+;
      re(i,,P*M)insert(S,i,,);
        re(i,,N)insert(P*M+i,T,p[i],);
        re(i,,N)re(j,,M)insert((j-)*P+,P*M+i,,t[i][j]);
      cout<<mincostmaxflow()<<endl;
      return ;
  }