Dining
Description
Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.
Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.
Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and prepared D (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink
a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.
Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).
Input
Line 1: Three space-separated integers: N, F, and D
Lines 2..N+1: Each line i starts with a two integers Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fi integers
denote the dishes that cow i will eat, and the Di integers following that denote the drinks that cow i will drink.
Output
Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes
Sample Input
4 3 3
2 2 1 2 3 1
2 2 2 3 1 2
2 2 1 3 1 2
2 1 1 3 3
Sample Output
3
Hint
One way to satisfy three cows is:
Cow 1: no meal
Cow 2: Food #2, Drink #2
Cow 3: Food #1, Drink #1
Cow 4: Food #3, Drink #3
The pigeon-hole principle tells us we can do no better since there are only three kinds of food or drink. Other test data sets are more challenging, of course.——————————————————————————————————————题目的意思是给牛喂食物和饮料，每头牛有各自的喜爱的食物和饮料，只有当它同时拿到喜爱的食物和饮料他会满足，问最多多少头牛满足。思路：网络最大流。 我们可以把食物和牛连边流量为1，饮料和牛连边流量为1，为了防止每头牛得到多个食物和饮料，我们可以把一头牛拆成2个点，再加上一个源点和一个汇点，源点和所有食物连边流量为1，汇点和所有饮料连边流量为1，这样计算源点到汇点的最大流就可以了。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>using namespace std;#define LL long long
const int INF = 0x3f3f3f3f;
#define MAXN 500struct node
{
int u, v, next, cap;
} edge[MAXN*MAXN];
int nt[MAXN], s[MAXN], d[MAXN], visit[MAXN];
int cnt;void init()
{
cnt = 0;
memset(s, -1, sizeof(s));
}void add(int u, int v, int c)
{
edge[cnt].u = u;
edge[cnt].v = v;
edge[cnt].cap = c;
edge[cnt].next = s[u];
s[u] = cnt++;
edge[cnt].u = v;
edge[cnt].v = u;
edge[cnt].cap = 0;
edge[cnt].next = s[v];
s[v] = cnt++;
}bool BFS(int ss, int ee)
{
memset(d, 0, sizeof d);
d[ss] = 1;
queue<int>q;
q.push(ss);
while (!q.empty())
{
int pre = q.front();
q.pop();
for (int i = s[pre]; ~i; i = edge[i].next)
{
int v = edge[i].v;
if (edge[i].cap > 0 && !d[v])
{
d[v] = d[pre] + 1;
q.push(v);
}
}
}
return d[ee];
}int DFS(int x, int exp, int ee)
{
if (x == ee||!exp) return exp;
int temp,flow=0;
for (int i = nt[x]; ~i ; i = edge[i].next, nt[x] = i)
{
int v = edge[i].v;
if (d[v] == d[x] + 1&&(temp = (DFS(v, min(exp, edge[i].cap), ee))) > 0)
{
edge[i].cap -= temp;
edge[i ^ 1].cap += temp;
flow += temp;
exp -= temp;
if (!exp) break;
}
}
if (!flow) d[x] = 0;
return flow;
}int Dinic_flow(int ss, int ee)
{
int ans = 0;
while (BFS(ss, ee))
{
for (int i = 0; i <= ee; i++) nt[i] = s[i];
ans+= DFS(ss, INF, ee);
}
return ans;
}int main()
{
int n,f,d,a,b,x;//f:1~f,牛:f+1~f+n&&f+n+1~f+2*n,d:f+2*n+1~f+2*n+d
while(~scanf("%d%d%d",&n,&f,&d))
{
init();
for(int i=0; i<n; i++)
{
scanf("%d%d",&a,&b);
for(int j=0; j<a; j++)
{
scanf("%d",&x);
add(x,f+1+i,1);
}
for(int j=0; j<b; j++)
{
scanf("%d",&x);
add(f+n+1+i,f+2*n+x,1);
}
}
for(int i=0; i<f; i++)
{
add(0,i+1,1);
}
for(int i=0; i<d; i++)
{
add(f+2*n+i+1,f+d+2*n+1,1);
}
for(int i=0; i<n; i++)
{
add(f+i+1,f+i+1+n,1);
}
printf("%d\n",Dinic_flow(0,f+d+2*n+1));
}
return 0;
}
