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POJ1149 PIGS [最大流 建图]
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.
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 33 1 102 1 2 22 1 3 31 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 MBSubmit: 183 Solved: 123
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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());}
POJ1149 PIGS [最大流 建图]
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