POJ 3281 —— 最大流

Dining

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7987		Accepted: 3644

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 F_i and D_i, the number of dishes that cow i likes and the number of drinks that cow i likes. The next F_i integers denote the dishes that cow i will eat, and the D_i 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.

Source

USACO 2007 Open Gold

題意是有N頭牛，他們每個有一些喜歡的食物和喜歡的飲料，問你最多有多少頭牛可以同時得到自己喜歡的飲料和食物。

思路是建一個超級源點s，一個超級匯點t，s指向所有的食物，食物指向喜歡它的牛，每頭牛結點拆成兩個，中間有一條邊相連，第二列的牛指向他喜歡的飲料，飲料結點都指向t，所有邊權都爲1，然後求s到t的最大流即可。（圖的規模比較小，用Ford-Fulkerson算法）

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 120 + 5;
const int MAXS = 10000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
const int inf = 1 << 30;
#define eps 1e-10
const long long MOD = 1000000000 + 7;
const int mod = 10007;
typedef long long LL;
const double PI = acos(-1.0);
//typedef double D;
typedef pair<int , int> pii;
typedef vector<int> vec;
typedef vector<vec> mat;
const int MAX_F = 120;
const int MAX_D = 120;

#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000";
int N , F , D;
bool likeF[MAXN][MAX_F];
bool likeD[MAXN][MAX_D];
struct edge{int to , cap , rev;};
vector<edge>G[MAXN << 2];
bool used[MAXN << 2];
void add_edge(int from , int to , int cap)
{
    G[from].push_back((edge){to , cap , G[to].size()});
    G[to].push_back((edge){from , 0 , G[from].size() - 1});
}
int dfs(int v , int t , int f)
{
    if(v == t)return f;
    used[v] = true;
    for(int i = 0 ; i < G[v].size() ; i++)
    {
        edge & e = G[v][i];
        if(!used[e.to] && e.cap > 0)
        {
            int d = dfs(e.to , t , min(f  , e.cap));
            if(d > 0)
            {
                e.cap -= d;
                G[e.to][e.rev].cap += d;
                return d;
            }

        }
    }
    return 0;
}
int max_flow(int s , int t)
{
    int flow  = 0;
    for(;;)
    {
        memset(used , 0 , sizeof(used));
        int f = dfs(s , t , INF);
        if(f == 0)return flow;
        flow += f;
    }
}
void solve()
{
    int s = N *2 + F + D;
    int t = s + 1;
    for(int i = 0 ; i < F ; i++)
    {
        add_edge(s , N * 2 + i, 1);
    }
    for(int i =0  ; i < D ; i++)
    {
        add_edge(N * 2 + F +i , t , 1);
    }
    for(int i = 0 ; i < N ; i++)
    {
        add_edge(i , N + i , 1);
        for(int j = 0 ; j < F ; j++)
        {
            if(likeF[i][j])add_edge(N * 2 + j , i , 1);
        }
        for(int j = 0 ; j < D ; j++)
        {
            if(likeD[i][j])add_edge(N + i , N * 2 + F + j , 1);
        }
    }
    printf("%d\n" , max_flow(s , t));
}
int main()
{
    while(~scanf("%d%d%d" , &N , &F, &D))
    {
        int x  , y;
        clr(likeD , 0);
        clr(likeF , 0);
        for(int i = 0 ; i < N ; i++)
        {
            scanf("%d%d" , &x , &y);
            while(x--)
            {
                int j;
                scanf("%d" ,&j);
                likeF[i][j - 1] = 1;
            }
            while(y--)
            {
                int j;
                scanf("%d" , &j);
                likeD[i][j - 1] = 1;
            }
        }
        solve();
    }
    return 0;
}