POJ 3281-Dining（最大流）

思路：主要是如何建图，因为每头牛只能喂一次所以要拆点，我们吧图分成5部分 超级源点——>食物——>牛——>牛（拆点后每头牛连自己的内个点）——>水——>超级汇点。跑一遍最大流

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
#include<queue>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e9+7;
const int N=2e6+10;
const int inf=0x7f7f7f7f;ll gcd(ll a,ll b)
{return b==0?a:gcd(b,a%b);
}ll lcm(ll a,ll b)
{return a*(b/gcd(a,b));
}template <class T>
void read(T &x)
{char c;bool op = 0;while(c = getchar(), c < '0' || c > '9')if(c == '-')op = 1;x = c - '0';while(c = getchar(), c >= '0' && c <= '9')x = x * 10 + c - '0';if(op)x = -x;
}
template <class T>
void write(T x)
{if(x < 0)x = -x, putchar('-');if(x >= 10)write(x / 10);putchar('0' + x % 10);
}int  s,t;
int ans,dis[N];
int cnt=1,now[N],head[N];
struct node{int to,nex;ll cap;
}edge[N];
void add(int u,int v,ll w)
{edge[++cnt].to=v;edge[cnt].cap=w;edge[cnt].nex=head[u];head[u]=cnt;edge[++cnt].to=u;edge[cnt].cap=0;edge[cnt].nex=head[v];head[v]=cnt;
}bool bfs()
{for(int i=1;i<=10086;i++) dis[i]=inf;queue<int> q;q.push(s);dis[s]=0;now[s]=head[s];while(!q.empty()){int x=q.front();q.pop();for(int i=head[x];i;i=edge[i].nex){int v=edge[i].to;if(edge[i].cap>0&&dis[v]==inf){q.push(v);now[v]=head[v];dis[v]=dis[x]+1;if(v==t) return true;}}}return false;
}int dfs(int x,ll sum)
{if(x==t) return sum;int k,res=0;for(int i=now[x];i&&sum;i=edge[i].nex){now[x]=i;int v=edge[i].to;if(edge[i].cap>0&&(dis[v]==dis[x]+1)){k=dfs(v,min(sum,edge[i].cap));if(k==0) dis[v]=inf;edge[i].cap-=k;edge[i^1].cap+=k;res+=k;sum-=k;}}return res;
}
int dinin()
{ans=0;while(bfs()){//cout<<2333<<endl;ans+=dfs(s,inf);}return ans;
}int main()
{s=0,t=505;int f,d,n;while(~scanf("%d%d%d",&n,&f,&d)){memset(head,0,sizeof head);cnt=1;for(int i=1;i<=f;i++){add(s,i,1);}for(int i=1;i<=d;i++){add(i+300,t,1);}for(int i=1;i<=n;i++){add(i+100,i+200,1);int F,D;scanf("%d%d",&F,&D);for(int j=1;j<=F;j++){int x;scanf("%d",&x);add(x,i+100,1);}for(int j=1;j<=D;j++){int x;scanf("%d",&x);add(i+200,x+300,1);}}int res=dinin();printf("%d\n",ans);}return 0;
}