Farmer John completed his new barn just last week,
complete with all the latest milking technology. Unfortunately, due to
engineering problems, all the stalls in the new barn are different. For the
first week, Farmer John randomly assigned cows to stalls, but it quickly became
clear that any given cow was only willing to produce milk in certain stalls.
For the last week, Farmer John has been collecting data on which cows are
willing to produce milk in which stalls. A stall may be only assigned to one
cow, and, of course, a cow may be only assigned to one stall.
Given the preferences of the cows, compute the maximum
number of milk-producing assignments of cows to stalls that is possible.
Input. The input includes several cases. For each case, the
first line contains two integers, N (0 ≤ N ≤ 200) and M (0 ≤
M ≤ 200). N is the number of cows that Farmer John has and M is the
number of stalls in the new barn. Each of the following N lines corresponds to
a single cow. The first integer (Si) on the line is the number of stalls that
the cow is willing to produce milk in (0 ≤ Si ≤ M). The subsequent
Si integers on that line are the stalls in which that cow is willing to produce
milk. The stall numbers will be integers in the range (1..M), and no stall will
be listed twice for a given cow.
Output. For each case, output a single line with a single
integer, the maximum number of milk-producing stall assignments that can be
made.
Sample Input
5 5
2 2 5
3 2 3 4
2 1 5
3 1 2 5
1 2
Sample
Output
4
графы – максимальное паросочетание
Ответ равен максимальному
паросочетанию в двудольном графе, образованном коровами и стойлами. Желания
коров производить молоко в определенных стойлах задают ребра двудольного графа.
Реализация алгоритма
#include <cstdio>
#include <vector>
#define MAX
210
using namespace std;
vector<vector<int> > g;
vector<int> used, mt, par;
int i, j,
ptr;
int n, m,
a, b, k, flow;
int dfs(int v)
{
if (used[v]) return 0;
used[v] = 1;
for (int i = 0; i < g[v].size(); i++)
{
int to =
g[v][i];
if (mt[to]
== -1 || dfs(mt[to]))
{
mt[to] = v;
par[v] = 1;
return 1;
}
}
return 0;
}
void
AugmentingPath(void)
{
int i, run;
mt.clear();
mt.assign (m+1, -1);
par.clear(); par.assign (n+1, -1);
for (run = 1;
run; )
{
run = 0; used.assign(n+1, 0);
for (i = 1;
i <= n; i++)
if
((par[i] == -1) && dfs(i)) run = 1;
}
}
int main(void)
{
while(scanf("%d %d",&n,&m) == 2)
{
g.clear();
g.resize(n+1);
for(i = 1;
i <= n; i++)
{
scanf("%d",&k);
while(k--)
{
scanf("%d",&b);
g[i].push_back(b);
}
}
AugmentingPath();
for (flow =
0, i = 1; i <= m; i++)
if (mt[i]
!= -1) flow++;
printf("%d\n",flow);
}
return 0;
}