PIGS
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 20662 | Accepted: 9435 |
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
中文题面
1280: Emmy卖猪pigs
Time Limit: 1 Sec Memory Limit: 162 MB
Submit: 183 Solved: 123
[Submit][Status][Discuss]
Description
Emmy在一个养猪场工作。这个养猪场有M个锁着的猪圈，但Emmy并没有钥匙。顾客会到养猪场来买猪，一个接着一个。每一位顾客都会有一些猪圈的钥匙，他们会将这些猪圈打开并买走固定数目的猪。 所有顾客有的钥匙和他们需要买猪的数量在事先都告诉了Emmy，于是Emmy要订一个计划，使得卖出去的猪最多。 买卖的过程是这样的：一个顾客前来，并打开所有他可以打开的猪圈。然后Emmy从这些猪圈里牵出固定数目的猪卖给顾客（最多只能和顾客需要数相等），并可以重新安排这些开着的猪圈中的猪。 每个猪圈可以存放任意数目的猪。 写一个程序，使得Emmy能够卖出去尽可能多的猪。
Input
第一行有两个整数：M和N，表示猪圈数和顾客数。 第二行有M个整数，表示每个猪圈初始时有多少猪。 接下来的N行按照前来的次序描述了每一个顾客，每行的格式如下： A K1 K2…KA B A表示该顾客拥有的钥匙数，K1...KA表示每个钥匙所对应的猪圈，B表示该顾客需要购买的猪的数目。
Output
仅包含一个整数，即最多能卖出去的猪的数目。
朴素见图的话，因为一个人的购买影响下一个人，所以可以按每个购买分层
猪圈和人作为点，s连猪圈一开始数量，人连t购买数
每个人(购买)作为一个层次，从上个层次到下个层次同一个猪圈连INF，然后可以购买的(能合并在一起)互相连起来、
这样点n+nm，边2nm
考虑一些边没用，没必要每个人的购买都重新弄一批猪圈的点，保存每个猪圈当前到了那个人手里然后连INF就行了，因为下一个人能买这个猪圈，以前拿着猪圈的人打开的所有猪圈都可以
这样点n，边nm
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=105,M=1005,INF=1e9;
inline int read(){
char c=getchar();int x=0,f=1;
while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();}
while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();}
return x*f;
}
int m,n,s,t;
int pig[M],now[M];
struct edge{
int v,c,f,ne;
}e[N*M<<1];
int cnt,h[N];
inline void ins(int u,int v,int c){
cnt++;
e[cnt].v=v;e[cnt].c=c;e[cnt].f=0;e[cnt].ne=h[u];h[u]=cnt;
cnt++;
e[cnt].v=u;e[cnt].c=0;e[cnt].f=0;e[cnt].ne=h[v];h[v]=cnt;
}
int q[N],head,tail,vis[N],d[N];
bool bfs(){
memset(vis,0,sizeof(vis));
memset(d,0,sizeof(d));
head=tail=1;
d[s]=0;vis[s]=1;
q[tail++]=s;
while(head!=tail){
int u=q[head++];
for(int i=h[u];i;i=e[i].ne){
int v=e[i].v;
if(!vis[v]&&e[i].c>e[i].f){
vis[v]=1;
d[v]=d[u]+1;
q[tail++]=v;
if(v==t) return true;
}
}
}
return false;
}
int cur[N];
int dfs(int u,int a){
if(u==t||a==0) return a;
int flow=0,f;
for(int &i=cur[u];i;i=e[i].ne){
int v=e[i].v;
if(d[v]==d[u]+1&&(f=dfs(v,min(a,e[i].c-e[i].f)))>0){
flow+=f;
e[i].f+=f;
e[((i-1)^1)+1].f-=f;
a-=f;
if(a==0) break;
}
}
return flow;
}
int dinic(){
int flow=0;
while(bfs()){
for(int i=s;i<=t;i++) cur[i]=h[i];
flow+=dfs(s,INF);
}
return flow;
}
int main(){
//freopen("in.txt","r",stdin);
m=read();n=read();s=0;t=n+1;
for(int i=1;i<=m;i++) pig[i]=read();
for(int i=1;i<=n;i++){
int A=read(),B,x;
while(A--){
x=read();
if(!now[x]) ins(s,i,pig[x]),now[x]=i;
else ins(now[x],i,INF),now[x]=i;
}
B=read();
ins(i,t,B);
}
printf("%d",dinic());
}