POJ 1149 PIGS

Posted 2020-09-02 り挽歌、花开花落的流年

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20892		Accepted: 9549

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 3
3 1 10
2 1 2 2
2 1 3 3
1 2 6

Sample Output

7

Source

Croatia OI 2002 Final Exam - First day

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<queue>
 5 #include<vector>
 6 #define maxn 1050
 7 using namespace std;
 8 int dep[110],m,n,pig[maxn],house[maxn];
 9 int vis[110],map[110][110];
10 #define INF 0x7fffffff
11 bool BFS(){
12     memset(dep,-1,sizeof(dep));
13     queue<int>q;
14     dep[0]=0;q.push(0);
15     while(!q.empty()){
16         int u=q.front();q.pop();
17         for(int v=1;v<=n+1;v++){
18             if(map[u][v]>0&&dep[v]==-1){
19                 dep[v]=dep[u]+1;
20                 if(v==n+1)return 1;
21                 else q.push(v);
22             }
23         }
24     }
25     return 0;
26 }
27 int Dinic(){
28     int maxf=0;vector<int>q;
29     while(BFS()){
30         memset(vis,0,sizeof(vis));
31         q.push_back(0);vis[0]=1;
32         while(!q.empty()){
33             int p=q.back();
34             if(p==n+1){
35                 int minn,minx=0x7fffffff;
36                 for(int i=1;i<q.size();i++){
37                     int u=q[i-1],v=q[i];
38                     if(map[u][v]<minx){
39                         minx=map[u][v];
40                         minn=u;
41                     }
42                 }
43                 maxf+=minx;
44                 for(int i=1;i<q.size();i++){
45                     int u=q[i-1],v=q[i];
46                     map[u][v]-=minx;map[v][u]+=minx;
47                 }
48                 while(!q.empty()&&q.back()!=minn){
49                     vis[q.back()]=0;q.pop_back();
50                 }
51             }
52             else {
53                 int i;
54                 for(i=0;i<=n+1;i++){
55                     if(map[p][i]>0&&dep[i]==dep[p]+1
56                         &&!vis[i]){
57                         q.push_back(i);vis[i]=1;
58                         break;
59                     }
60                 }
61                 if(i>n+1)q.pop_back();
62             }
63         }
64     }
65     return maxf;
66 }
67 int main(){
68     cin>>m>>n;
69     for(int i=1;i<=m;i++)
70       scanf("%d",pig+i);
71     
72     int num,k;
73     for(int i=1;i<=n;i++){
74         scanf("%d",&num);
75         for(int j=0;j<num;j++){
76             scanf("%d",&k);
77             if(house[k]==0) map[0][i]+=pig[k];
78             else map[house[k]][i]=INF;
79             house[k]=i;
80         }
81         int tpig;
82         scanf("%d",&tpig);
83         map[i][n+1]=tpig;
84     }
85     cout<<Dinic()<<endl;
86     return 0;
87 }