POJ 1149 PIGS
PIGS
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 20582 | Accepted: 9389 |
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
题意：
有m个猪圈，n个人，每个人可以打开一些猪圈，从这些猪圈里领猪，当前开着的猪圈中的猪可以随意走动到开着的猪圈中，每个客人走之后猪圈会再次关上，每个客人有期望数量，问最多可以领走多少小猪^(*￣(oo)￣)^
分析：
“这水题写什么写”---来自YSQ的嘲讽TAT...
确实没什么好写的...
不过有一点思想很重要就是可以用+∞的边表示流量传递...
对于每个猪圈我们从S向每个猪圈连一条容量为猪数量的边，从每个顾客到T连一条容量为期望领养数量的边，对于每一个客人，如果这个客人是第一次打开这个猪圈的客人，那么猪圈向这个客人连+∞的边，否则上一个打开这个猪圈的客人向当前客人连一条+∞的边...
代码：
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 //by NeighThorn 6 #define inf 0x3f3f3f3f 7 using namespace std; 8 //大鹏一日同风起，扶摇直上九万里 9 10 const int maxn=2000+5,maxm=800000+5; 11 12 int n,m,S,T,cnt,hd[maxn],to[maxm],fl[maxm],nxt[maxm],pre[maxn],pos[maxn],num[maxn]; 13 14 inline void add(int s,int x,int y){ 15 fl[cnt]=s;to[cnt]=y;nxt[cnt]=hd[x];hd[x]=cnt++; 16 fl[cnt]=0;to[cnt]=x;nxt[cnt]=hd[y];hd[y]=cnt++; 17 } 18 19 inline bool bfs(void){ 20 memset(pos,-1,sizeof(pos)); 21 int head=0,tail=0,q[maxn]; 22 q[0]=S;pos[S]=0; 23 while(head<=tail){ 24 int top=q[head++]; 25 for(int i=hd[top];i!=-1;i=nxt[i]) 26 if(pos[to[i]]==-1&&fl[i]) 27 pos[to[i]]=pos[top]+1,q[++tail]=to[i]; 28 } 29 return pos[T]!=-1; 30 } 31 32 inline int find(int v,int f){ 33 if(v==T) 34 return f; 35 int res=0,t; 36 for(int i=hd[v];i!=-1&&f>res;i=nxt[i]) 37 if(pos[to[i]]==pos[v]+1&&fl[i]) 38 t=find(to[i],min(fl[i],f-res)),res+=t,fl[i]-=t,fl[i^1]+=t; 39 if(!res) 40 pos[v]=-1; 41 return res; 42 } 43 44 inline int dinic(void){ 45 int res=0,t; 46 while(bfs()) 47 while(t=find(S,inf)) 48 res+=t; 49 return res; 50 } 51 52 signed main(void){ 53 scanf("%d%d",&m,&n);cnt=0; 54 S=0,T=n+m+1;memset(hd,-1,sizeof(hd)); 55 for(int i=1;i<=m;i++) 56 scanf("%d",&num[i]),pre[i]=i,add(num[i],S,i); 57 for(int i=1;i<=n;i++){ 58 int x;scanf("%d",&x); 59 for(int j=1,y;j<=x;j++) 60 scanf("%d",&y),add(inf,pre[y],i+m),pre[y]=i+m; 61 scanf("%d",&x);add(x,i+m,T); 62 } 63 printf("%d\n",dinic()); 64 return 0; 65 }//Cap ou pas cap. Cap;
By NeighThorn
