PIGSTime Limit:1000MSMemory Limit:10000KTotal Submissions:19068Accepted:8697
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 33 1 102 1 2 22 1 3 31 2 6
Sample Output
7
Source
Croatia OI 2002 Final Exam - First day最大流，
poj1149 PIGS
，電腦資料
《poj1149 PIGS》(http://www.shangyepx.com)。構圖方式：
①把每個顧客看作除源點和匯點以外的節(jié)點。
②從源點向每個豬圈的第一個顧客連一條邊，容量為該豬圈最初的豬的數(shù)量。
③每個豬圈的前后兩個顧客之間連一條邊，容量為正無窮。因為可以任意分配每個豬圈中的豬的數(shù)量。
④從每個顧客向匯點連一條邊，容量為要購買的豬的數(shù)量。
這道題的構圖方法很巧妙。
#include<iostream>#include<cstdio>#include#include<cmath>#include<cstring>#include<cstdlib>#include<queue>#define F(i,j,n) for(int i=j;i<=n;i++)#define D(i,j,n) for(int i=j;i>=n;i--)#define LL long long#define pa pair<int,int>#define MAXN 105#define MAXM 1005#define INF 1000000000using namespace std;int n,m,k,x,s,t,cnt=1,ans=0;int pre[MAXM],head[MAXN],cur[MAXN],dis[MAXN],c[MAXN],a[MAXM];bool vst[MAXM],f[MAXN];struct edge_type{ int next,to,v;}e[10005];inline int read(){ int x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}inline void add_edge(int x,int y,int v){ e[++cnt]=(edge_type){head[x],y,v};head[x]=cnt; e[++cnt]=(edge_type){head[y],x,0};head[y]=cnt;}inline bool bfs(){ queue<int>q; memset(dis,-1,sizeof(dis)); dis[s]=0;q.push(s); while (!q.empty()) { int tmp=q.front();q.pop(); if (tmp==t) return true; for(int i=head[tmp];i;i=e[i].next) if (e[i].v&&dis[e[i].to]==-1) { dis[e[i].to]=dis[tmp]+1; q.push(e[i].to); } } return false;}inline int dfs(int x,int f){ int tmp,sum=0; if (x==t) return f; for(int &i=cur[x];i;i=e[i].next) { int y=e[i].to; if (e[i].v&&dis[y]==dis[x]+1) { tmp=dfs(y,min(f-sum,e[i].v)); e[i].v-=tmp;e[i^1].v+=tmp;sum+=tmp; if (sum==f) return sum; } } if (!sum) dis[x]=-1; return sum;}inline void dinic(){ while (bfs()) { F(i,1,n+2) cur[i]=head[i]; ans+=dfs(s,INF); }}int main(){ memset(head,0,sizeof(head)); memset(c,0,sizeof(c)); memset(vst,false,sizeof(vst)); m=read();n=read();s=n+1;t=n+2; F(i,1,m) a[i]=read(); F(i,1,n) { memset(f,false,sizeof(f)); k=read(); while (k--) { x=read(); if (!vst[x]) {vst[x]=true;c[i]+=a[x];} if (pre[x]&&!f[pre[x]]) add_edge(pre[x],i,INF),f[pre[x]]=true; pre[x]=i; } x=read();if (x) add_edge(i,t,x); } F(i,1,n) if (c[i]) add_edge(s,i,c[i]); dinic(); printf("%d\n",ans);}</int></int,int></queue></cstdlib></cstring></cmath></cstdio></iostream>