Poj 3281 dining network stream

Nheaded ox, d kind of beverage, F kind of food, every day the ox eats one kind of food one kind of beverage, both food and drink have only one. Ask the maximum number of cows.

Diagram: as each ox only needs one drink and one food, each Ox needs to be split into two points, with a capacity of 1. From the start point to the side with a capacity of 1 for all foods, and from the beverage to the sink point, the side with a capacity of 1. The ox connects with food and drinks.

Just take a closer look at the graph ..

# Include <cstdio>
# Include <iostream>
# Include <cstring>
# Include <algorithm>
# Define n 5005
# Define M 100005 // This should be large enough ..
# Define INF 999999999
Using namespace STD;

Int S, T, num, adj [N], DIS [N], Q [N];
Struct edge
{
Int V, W, pre;
} E [m];

Int min (INT g, int H)
{
If (G> H)
Return h;
Else
Return g;
}
Void insert (int u, int V, int W) // The Insert above can be written as follows ..
{
E [num]. V = V;
E [num]. W = W;
E [num]. Pre = adj [u];
Adj [u] = num ++;
E [num]. V = u;
E [num]. W = 0;
E [num]. Pre = adj [v];
Adj [v] = num ++;

}
Int BFS ()
{
Int I, X, V, tail = 0, head = 0;
Memset (DIS, 0, sizeof (DIS ));
Dis [s] = 1;
Q [tail ++] = s;
While (Head <tail)
{
X = Q [head ++];
For (I = adj [X]; I! =-1; I = E [I]. PRE)
If (E [I]. W & dis [V = E [I]. V] = 0)
{
Dis [v] = dis [x] + 1;
If (V = T)
Return 1;
Q [tail ++] = V;
}
}
Return 0;
}
Int DFS (int s, int limit)
{
If (S = T)
Return limit;
Int I, V, TMP, cost = 0;
For (I = adj [s]; I! =-1; I = E [I]. PRE)
If (E [I]. W & dis [s] = dis [V = E [I]. V]-1)
{
TMP = DFS (v, min (Limit-cost, E [I]. W ));
If (TMP> 0)
{
E [I]. W-= TMP;
E [I ^ 1]. W + = TMP;
Cost + = TMP;
If (Limit = cost)
Break;
}
Else dis [v] =-1;
}
Return cost;
}
Int dinic ()
{
Int ans = 0;
While (BFS ())
Ans + = DFS (S, INF );
Return ans;
}
Int main ()
{
Int num, F, D;
Int I, j, FJ, DJ, F, D;
While (scanf ("% d", & num, & F, & D )! = EOF)
{
Memset (adj,-1, sizeof (adj ));
Num = 0;
For (I = 1; I <= f; I ++)
Insert (0, I, 1 );
For (I = F + 2 * num + 1; I <= F + 2 * num + D; I ++)
Insert (I, F + D + 2 * num + 1, 1 );
For (I = F + 1; I <= F + num; I ++)
Insert (I, I + num, 1 );
For (I = 1; I <= num; I ++)
{
Cin> F> D;
For (j = 1; j <= f; j ++)
{
Cin> FJ;
Insert (FJ, I + F, 1 );
}
For (j = 1; j <= D; j ++)
{
Cin> DJ;
Insert (F + num + I, F + 2 * num + DJ, 1 );
}
}
S = 0;
T = F + D + 2 * num + 1;
Printf ("% d \ n", dinic ());
}
Return 0;
}