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
題意:每個人想買豬2333,然後由m個豬圈,一個人會挑固定幾個豬圈打開,然後從裏面挑豬,被打開的豬圈可以隨便改變豬的個數,只要總和不變即可.
建圖思路:把每個顧客看成節點,將源點會第一個打開豬圈的顧客連起來,權值是顧客會第一次打開的豬圈中豬的總和,後面打開豬圈的顧客可挑選的豬相當於前一個顧客挑剩的總和.所以把前一個打開豬圈的人和當前打開的人連一條INF的邊,最後所有人和匯點連一條邊就好了 23333 建圖還是巧妙啊QWQ
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<queue>
#include<map>
#include<sstream>
#include<stack>
#include<functional>
#include<cctype>
using namespace std;
typedef long long ll;
//thanks to pyf ...
//thanks to lmd ...
#define INF 0x3f3f3f3f
#define CLR(a,b) memset(a,b,sizeof(a))
const int N = 5005;
struct Edge
{
int from,to,cap,flow;
Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct Dinic
{
int n,m,s,t;
vector<Edge>edges;
vector<int>G[N];
bool vis[N];
int d[N];
int cur[N];
void add_edge(int u,int v,int cap)
{
edges.push_back(Edge(u,v,cap,0));
edges.push_back(Edge(v,u,0,0));
int m = edges.size();
G[u].push_back(m-2);
G[v].push_back(m-1);
}
bool bfs()
{
memset(vis,0,sizeof(vis));
queue<int>q;
q.push(s);
d[s] = 0;
vis[s] = 1;
while(!q.empty())
{
int x = q.front();
q.pop();
for(int i=0;i<G[x].size();i++)
{
Edge &e = edges[G[x][i]];
if(!vis[e.to]&&e.cap > e.flow)
{
vis[e.to] = 1;
d[e.to] = d[x] +1;
q.push(e.to);
}
}
}
return vis[t];
}
int dfs(int x,int a)
{
if(x==t||a==0)
return a;
int flow = 0,f;
for(int &i = cur[x];i<G[x].size();i++)
{
Edge &e = edges[G[x][i]];
if(d[x]+1==d[e.to]&&(f=dfs(e.to,min(a,e.cap-e.flow)))>0)
{
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a-=f;
if(!a)
break;
}
}
return flow;
}
int Maxflow(int s,int t)
{
this -> s = s,this -> t = t;
int flow = 0;
while(bfs())
{
CLR(cur,0);
flow+= dfs(s,INF);
}
return flow;
}
};
int pig[N];
int vis[N];
int used[N];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)==2)
{
Dinic ans;
int s = 0;
int t = m+1;
CLR(vis,0);
for(int i=1;i<=n;i++)
{
scanf("%d",pig+i);
}
for(int i=1;i<=m;i++)
{
int cnt;
int cap = 0;
CLR(used,0);
scanf("%d",&cnt);
for(int j=0;j<cnt;j++)
{
int id;
scanf("%d",&id);
if(!vis[id])
{
vis[id] = i;
cap+=pig[id];
}
else
{
if(!used[vis[id]])
ans.add_edge(vis[id],i,INF);
vis[id] = i;
}
}
if(cap)
ans.add_edge(s,i,cap);
int need;
scanf("%d",&need);
ans.add_edge(i,t,need);
}
printf("%d\n",ans.Maxflow(s,t));
}
}