题目链接:POJ3281 Dining
题意:f个食物和d个饮料,分别有自己的编号,n个奶牛吃到自己想要的食物和饮料才会开心,求最多多少奶牛开心;
分析:对奶牛而言有1的上限,对奶牛拆点建边,流量为1(注意点的编号);
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
#include<string.h>
#include<cstring>
#define pb push_back
using namespace std;
typedef long long ll;
const int maxn=1200;
const int mod=1e9+7;
const int inf=0x3f3f3f3f;
struct node {int to,cap,rev;};
int dep[maxn],n,m,S,T;
vector<node>v[maxn];
void add(int from,int to,int cap) //重边情况不影响
{
v[from].pb({to,cap,(int)v[to].size()});
v[to].pb({from,0,(int)v[from].size()-1});
}
int bfs(int s,int t)
{
queue<int>q;
memset(dep,-1,sizeof(dep));
dep[s]=0;
q.push(s);
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=0;i<v[u].size();i++)
if(v[u][i].cap>0&&dep[v[u][i].to]==-1)
{
dep[v[u][i].to]=dep[u]+1;
q.push(v[u][i].to);
}
}
return dep[t]!=-1;
}
int dfs(int u,int t,int micap)
{
if(u==t)return micap;
int tmp;
for(int i=0;i<v[u].size();i++)
{
node &now=v[u][i];
if(now.cap>0 && dep[u]+1==dep[now.to] && (tmp=dfs(now.to,t,min(now.cap,micap))))
{
now.cap-=tmp;
v[now.to][now.rev].cap+=tmp;
return tmp;
}
}
return 0;
}
int dinic(int s,int t)
{
int ans=0,tmp;
while(bfs(s,t))
while(1)
{
tmp=dfs(s,t,inf);
if(tmp==0)break;
ans+=tmp;
}
return ans;
}
int main()
{
int n,f,d;scanf("%d%d%d",&n,&f,&d);
int s=0,t=2*n+f+d+1;memset(v,0,sizeof(v));
for(int i=1;i<=f;i++) add(s,i,1);
for(int i=1;i<=d;i++) add(i+f,t,1);
int res=f+d;
for(int i=1;i<=n;i++)
{
int x,y,kk;scanf("%d%d",&x,&y);add(i+res,i+res+n,1);
while(x--) {scanf("%d",&kk);add(kk,i+res,1);}
while(y--) {scanf("%d",&kk);add(i+res+n,kk+f,1);}
}
printf("%d\n",dinic(s,t));
}