Network stream (Max Stream): POJ 1149 PIGS

Pigstime limit:1000msmemory limit:10000kbthis problem'll be judged onPKU.
64-bit integer(integer) IO format: %lld Java class name: Main 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 dataconcerning(about)Customers planning to visit the farm on this particular day is available to Mirko early in the morning so that he can mak e a sales-plan in order toMaximize(Take ... Maximum value)The number of pigs sold.
Moreprecisely(precisely), theprocedure(program)is as following:the customer arives, opens all pig-houses to which he have the key, Mirko sells a certain number of pigs F Rom all the unlocked pig-houses to him, and, if Mirko wants, he canRedistribute(Redistribution)The remaining 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 ofinput(Input)Containsintegers(integer)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 (sortednondecreasingly(not reduced)) and that he wants to buy B pigs. Numbers A and B can be equal to 0. Output the first and only line of theOutput(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
　　
Modeling problem, you need to pay attention to the optimization of space.

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <queue>5 6 using namespacestd;7 Const intinf=2147483647;8 Const intmaxn=1010, maxm=4010;9 intCNT,FIR[MAXN],NXT[MAXM],CAP[MAXM],TO[MAXM],DIS[MAXN],GAP[MAXN],PATH[MAXN],USED[MAXN];Ten  One voidAddedge (intAintBintc) A { -nxt[++cnt]=Fir[a]; -to[cnt]=b; thecap[cnt]=C; -fir[a]=CNT; - } -  + BOOLBFS (intSintT) - { +memset (DIS,0,sizeof(DIS)); Adis[t]=1; atqueue<int>Q;q.push (T); -      while(!q.empty ()) -     { -         intNode=Q.front (); Q.pop (); -          for(intI=fir[node];i;i=Nxt[i]) -         { in             if(Dis[to[i]])Continue; -dis[to[i]]=dis[node]+1; to Q.push (To[i]); +         } -     } the     returnDis[s]; * } $ intFRON[MAXN];Panax Notoginseng intISAP (intSintT) - { the     if(!BFS (s,t)) +         return 0; A      for(intI=1; i<=t;i++) + +Gap[dis[i]]; the     intp=s,ret=0; +memcpy (Fron,fir,sizeof(FIR)); -      while(dis[s]<=T) $     { $         if(p==T) { -             intf=INF; -              while(p!=R) { thef=min (f,cap[path[p]]); -p=to[path[p]^1];Wuyi             } thep=t;ret+=F; -              while(p!=R) { Wucap[path[p]]-=F; -cap[path[p]^1]+=F; Aboutp=to[path[p]^1]; $             } -         } -         int&ii=Fron[p]; -          for(; ii;ii=Nxt[ii]) { A             if(!cap[ii]| | dis[to[ii]]+1!=Dis[p]) +                 Continue; the             Else  -                  Break; $         }         the         if(ii) { thep=To[ii]; thepath[p]=II; the         } -         Else{ in             if(--gap[dis[p]]==0) Break; the             intminn=t+1; the              for(intI=fir[p];i;i=Nxt[i]) About                 if(Cap[i]) theminn=min (Minn,dis[to[i]]); thegap[dis[p]=minn+1]++; thefron[p]=Fir[p]; +             if(p!=S) -p=to[path[p]^1];  the         }Bayi     } the     returnret; the } -  - voidInit () the { thememset (Fir,0,sizeof(FIR)); thememset (Used,0,sizeof(used));  theCnt=1; - } the intMain () the { the     intN,m,num,k,need;94      while(~SCANF ("%d%d",&m,&N)) the     { the Init (); the          for(intI=1; i<=m;i++){98scanf"%d",&num); AboutAddedge (0, i,num); -Addedge (I,0,0);101         }102          for(inti=m+1; i<=m+n;i++){103scanf"%d",&k);104              while(k--){ thescanf"%d",&num);106                 if(Used[num]) {107 Addedge (used[num],i,inf);108Addedge (I,used[num],0);109                 } the                 Else{111used[num]=i; the Addedge (num,i,inf);113Addedge (I,num,0); the                 } the                  the             }117scanf"%d",&need);118Addedge (i,n+m+1, need);119Addedge (n+m+1I0); -         }121printf"%d\n", ISAP (0, n+m+1));122     }123     return 0;124}

Network stream (Max Stream): POJ 1149 PIGS