| Time Limit: 1000MS | Memory Limit: 10000KB | 64bit IO Format: %I64d & %I64u |
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.
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.
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 3 3 1 10 2 1 2 2 2 1 3 3 1 2 6
Sample Output
7
Source
建图思路很好
此处搬运自AaronPolaris的博客:
构图方式：
①把每个顾客看作除源点和汇点以外的节点。
②从源点向每个猪圈的第一个顾客连一条边，容量为该猪圈最初的猪的数量。
③每个猪圈的前后两个顾客之间连一条边，容量为正无穷。因为可以任意分配每个猪圈中的猪的数量。
④从每个顾客向汇点连一条边，容量为要购买的猪的数量。
搬运完毕
1 //POJ - 1149 PIGS 2 /*by SilverN*/ 3 #include<iostream> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdio> 7 #include<cmath> 8 #include<queue> 9 using namespace std; 10 const int INF=0xfffff; 11 const int mxn=420;//最大顾客数 12 const int mxm=1200;//最大猪圈数 13 int s,t; 14 int w[mxn][mxn];//容量 15 int d[mxn];//深度 16 int house[mxm];//猪圈里猪数量 17 int last[mxn]; 18 int M,N;//猪圈数，顾客数 19 20 void init(){ 21 int i,j; 22 23 scanf("%d%d",&M,&N); 24 int num; 25 int x; 26 s=0;t=N+1; 27 for(i=1;i<=M;i++)scanf("%d",&house[i]); 28 for(i=1;i<=N;i++){ 29 scanf("%d",&num); 30 for(j=1;j<=num;j++){ 31 scanf("%d",&x); 32 if(!last[x])w[s][i]+=house[x]; 33 //如果是来到此猪圈的第一个顾客,从源点连边到顾客点，容量为猪圈里猪的数量 34 else w[last[x]][i]=INF; 35 //在同一个猪圈的前后顾客之间连边 36 last[x]=i; 37 } 38 scanf("%d",&x); 39 w[i][t]=x;//顾客到汇点连边，容量为欲购买量 40 } 41 return; 42 } 43 bool BFS(int s){ 44 queue<int>q; 45 memset(d,-1,sizeof(d)); 46 q.push(s); 47 d[s]=0; 48 int i; 49 while(!q.empty()){ 50 int u=q.front(); 51 if(u==t)return true; 52 q.pop(); 53 for(i=0;i<=t;i++){ 54 if(w[u][i] && d[i]==-1){ 55 d[i]=d[u]+1; 56 q.push(i); 57 } 58 } 59 } 60 return false; 61 } 62 int DFS(int x,int low){ 63 if(x==t)return low; 64 int i,a; 65 for(i=0;i<=t;i++){ 66 if(w[x][i]>0 && d[i]==d[x]+1) 67 if(a=DFS(i,min(w[x][i],low))){ 68 w[x][i]-=a; 69 w[i][x]+=a; 70 return a; 71 } 72 } 73 return 0; 74 } 75 void dinic(){ 76 int ans=0; 77 int flow=0; 78 while(BFS(s)){ 79 while(flow=DFS(s,INF)){ 80 ans+=flow; 81 } 82 } 83 printf("%d ",ans); 84 return; 85 } 86 int main(){ 87 init(); 88 dinic(); 89 return 0; 90 }