POJ 1149 PIGS

Posted 2020-07-26 SilverNebula

 

PIGS

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.

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

 

 

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\n",ans);
84     return;
85 }
86 int main(){
87     init();
88     dinic();
89     return 0;
90 }