POJ 1149 PIGS

PIGS

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %i64d &%i64u

Description

Mirko works on a pig farm the consists of M locked pig-houses and Mirko can ' t unlock any pighouse because he doesn ' t has The keys. Customers come to the farm one after another. Each of them have 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 is available to Mirko early in the mornin G So, he can make a sales-plan in order to maximize the number of pigs sold.
More precisely, the procedure was as following:the customer arrives, opens all pig-houses to which he had the key, Mirko s Ells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remain ing pigs across the unlocked pig-houses.
An unlimited number of pigs can is placed in every pig-house.
Write a program, that would find the maximum number of pigs that he can sell on this day.

Input

The first line of input contains integers m and N, 1 <= m <=, 1 <= N <=, number of pighouses and Number of customers. Pig houses is numbered from 1 to M and customers is numbered from 1 to N.
The next line contains M Integeres, for each pig-house initial number of pigs. The number of pigs in pig-house are 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 on the i-th customer is written in The (i+2)-th line):
A K1 K2 ... Ka B It means that this customer have 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

Source

Croatia OI 2002 Final Exam-first Day

The idea of building a map is very good

From Aaronpolaris's blog here:

Composition method:

① each customer as a node other than the source and sink points.

② an edge from the source point to the first customer of each pigsty, the volume of the first pig of the pigsty.

③ the front and back two customers of each pigsty have an edge, the capacity is positive infinity. Because the number of pigs in each pigsty can be arbitrarily assigned.

④ an edge from each customer to the meeting point, the capacity is the number of pigs to be purchased.

Transport complete

1 //POJ-1149 PIGS2 /*by Silvern*/3#include <iostream>4#include <algorithm>5#include <cstring>6#include <cstdio>7#include <cmath>8#include <queue>9 using namespacestd;Ten Const intinf=0xfffff; One Const intmxn=420;//Maximum number of customers A Const intmxm= -;//Maximum number of pigsty - ints,t; - intW[MXN][MXN];//capacity the intD[MXN];//Depth - intHOUSE[MXM];//Number of pigs in pigsty - intLAST[MXN]; - intM,n;//number of pigsty, number of customers +  - voidinit () { +     inti,j; A  atscanf"%d%d",&m,&N); -     intnum; -     intx; -s=0; t=n+1; -      for(i=1; i<=m;i++) scanf ("%d",&house[i]); -      for(i=1; i<=n;i++){ inscanf"%d",&num); -          for(j=1; j<=num;j++){ toscanf"%d",&x); +             if(!last[x]) w[s][i]+=House[x]; -             //if it is the first customer to come to this pigsty, from the source point of connection to the customer point, the capacity for the number of pigs in the pigsty the             Elsew[last[x]][i]=INF; *             //between the front and rear customers of the same pigsty $last[x]=i;Panax Notoginseng         } -scanf"%d",&x); theW[i][t]=x;//customers to the meeting point edge, capacity for the purchase amount +     } A     return; the } + BOOLBFS (ints) { -queue<int>Q; $memset (d,-1,sizeof(d)); $ Q.push (s); -d[s]=0; -     inti; the      while(!Q.empty ()) { -         intu=Q.front ();Wuyi         if(u==t)return true; the Q.pop (); -          for(i=0; i<=t;i++){ Wu             if(W[u][i] && d[i]==-1){ -d[i]=d[u]+1; About Q.push (i); $             } -         } -     } -     return false; A } + intDFS (intXintLow ) { the     if(x==t)returnLow ; -     intI,a; $      for(i=0; i<=t;i++){ the         if(w[x][i]>0&& d[i]==d[x]+1) the           if(a=DFS (I,min (W[x][i],low))) { thew[x][i]-=A; thew[i][x]+=A; -               returnA; in           } the     } the     return 0; About } the voidDinic () { the     intans=0; the     intflow=0; +      while(BFS (s)) { -          while(flow=DFS (S,inf)) { theans+=flow;Bayi         } the     } theprintf"%d\n", ans); -     return; - } the intMain () { the init (); the dinic (); the     return 0; -}

POJ 1149 PIGS