poj1149

PIGS

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 17919 Accepted: 8149

Description

Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs. 
All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold. 
More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses. 
An unlimited number of pigs can be placed in every pig-house. 
Write a program that will find the maximum number of pigs that he can sell on that day.

Input

The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N. 
The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000. 
The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line): 
A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.

Output

The first and only line of the output should contain the number of sold pigs.

Sample Input

3 33 1 102 1 2 22 1 3 31 2 6

Sample Output

7

题意： 有 M 个猪圈，每个猪圈里初始时有若干头猪。一开始所有猪圈都是关闭的。依次来了 N 个顾客，每个顾客分别会打开指定的几个猪圈，从中买若干头猪。每个顾客分别都有他能够买的数量的上限。每个顾客走后，他打开的那些猪圈中的猪，都可以被任意地调换到其它开着的猪圈里，然后所有猪圈重新关上。分析： 此题是网络流，网络流里经典的构图题。将顾客看作除源和汇以外的节点，源和每个猪圈的第一个顾客连边，边的权是开始时猪圈中猪的数目，若源和某个节点之间有重边，则将权合并，每个猪圈的前后相邻两顾客之间连边，由前一个顾客指向后一个顾客。每个顾客和汇之间连边，边的权是顾客所希望购买的猪的数目。网络流，需要对图进行化简化简规则：规律 1. 如果几个节点的流量的来源完全相同，且流量为+∞，则可以把它们合并成一个。规律 2. 如果几个节点的流量的去向完全相同，且流量为+∞，则可以把它们合并成一个。规律 3. 如果从点 u 到点 v 有一条容量为 +∞ 的边，并且 u 是 v 的唯一流量来源，或者 v 是 u 的唯一流量去向，则可以把 u 和 v 合并成一个节点。代码：

type  edge=record    y,r,next,op:longint;  end;var  g:array[0..10000] of edge;  level,q,h,pig,c,peo:array[0..10010] of longint;  cus:array [1..1010,1..1010] of longint;  n,m,f,i,j,a,b,tot,vs,ii,vt,x,y,k:longint;  ans:qword;  s:string;

function bfs:boolean;var  i,f,r,tmp,v,u:longint;begin  fillchar(level,sizeof(level),0);  f:=1;  r:=1;  q[f]:=vs;  level[vs]:=1;  repeat    v:=q[f];    tmp:=h[v];    while tmp<>-1 do      begin        u:=g[tmp].y;        if (g[tmp].r<>0) and (level[u]=0) then          begin            level[u]:=level[v]+1;            inc(r);    q[r]:=u;            if u=vt then exit(true);          end;        tmp:=g[tmp].next;      end;    inc(f);  until f>r;  exit(false);end;

function min(x,y:longint):longint;begin  if x<y then  exit(x);  exit(y);end;

function dfs(v,a:longint):longint;var  ans,flow,tmp,u,value:longint;begin  if (v=vt) or (a=0) then exit(a);  ans:=0;  tmp:=h[v];  while tmp<>-1 do    begin      u:=g[tmp].y;    value:=g[tmp].r;      if (level[u]=level[v]+1)  then        begin          flow:=dfs(u,min(a,value));          if flow<>0 then            begin              g[tmp].r:=g[tmp].r-flow;              g[g[tmp].op].r:=g[g[tmp].op].r+flow;              ans:=ans+flow;              a:=a-flow;              if a=0 then break;            end;        end;      tmp:=g[tmp].next;    end;  exit(ans);end;

procedure add(a,b,c:longint);begin  inc(tot);  g[tot].y:=b;  g[tot].r:=c;  g[tot].next:=h[a];  h[a]:=tot;  g[tot].op:=tot+1;  inc(tot);  g[tot].y:=a;  g[tot].r:=0;  g[tot].next:=h[b];  h[b]:=tot;  g[tot].op:=tot-1;end;begin    fillchar(h,sizeof(h),$ff);    readln(n,m);    tot:=0;    ans:=0;    vs:=0;    vt:=m+1;    for i:=1 to n do    read(pig[i]);    readln;    for i:=1 to m do    begin      read(x);      for j:=1 to x do      begin        read(y);        inc(c[y]);        cus[y,c[y]]:=i;      end;      read(y);      readln;       peo[i]:=y;    end;    for i:=1 to n do    begin      add(vs,cus[i,1],pig[i]);      for j:=2 to c[i] do      add(cus[i,j-1],cus[i,j],maxlongint);    end;    for i:=1 to m do    add(i,vt,peo[i]);    while bfs do    begin      ans:=ans+dfs(vs,maxlongint);    end;    writeln(ans);end.