poj1149 PIGS
PIGS
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 19068
Accepted: 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 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
最大流。
構圖方式：
①把每個顧客看作除源點和匯點以外的節點。
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量。
③每個豬圈的前後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量。
④從每個顧客向匯點連一條邊，容量為要購買的豬的數量。
這道題的構圖方法很巧妙。
#include
#include
#include
#include
#include
#include
#include
#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
#define MAXN 105
#define MAXM 1005
#define INF 1000000000
using 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()
{
queueq;
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);
}
