The Perfect Stall
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24081 | Accepted: 10695 |
Description
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
题目链接：POJ 1274
今天用做了这两道题才知道用最大流解决二分匹配的正确姿势 ：首先不要一开始就往S、T这两个点里加边，因为很可能一个点会多次与S相连，那这样这个点总的流量就不是1了。
假设二分图的左半部分是集合$L_i$、右半部分是集合$R_i$，那么首先从$L_i$到$R_i$连一条边，然后将$L_i$与$R_i$记录到各自的数字里，然后各自去重，再从S到去重后的左集合均连一条容量为1的边，从右集合$R_i$中连向T也是一条容量为1的边，然后再跑最大流，代码是POJ的题目代码，HDU的把左边也去重就好了。
代码：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=210<<1;
const int M=N*N*2;
struct edge
{
int to,nxt;
int cap;
};
edge E[M];
int head[N],tot;
int d[N];
vector<int>stall;
void init()
{
CLR(head,-1);
tot=0;
stall.clear();
}
void add(int s,int t,int c)
{
E[tot].to=t;
E[tot].cap=c;
E[tot].nxt=head[s];
head[s]=tot++;
E[tot].to=s;
E[tot].cap=0;
E[tot].nxt=head[t];
head[t]=tot++;
}
int bfs(int s,int t)
{
CLR(d,INF);
d[s]=0;
queue<int>Q;
Q.push(s);
while (!Q.empty())
{
int now=Q.front();
Q.pop();
for (int i=head[now]; ~i; i=E[i].nxt)
{
int v=E[i].to;
if(d[v]==INF&&E[i].cap)
{
d[v]=d[now]+1;
if(v==t)
return 1;
Q.push(v);
}
}
}
return d[t]!=INF;
}
int dfs(int s,int t,int f)
{
if(s==t||!f)
return f;
int ret=0;
for (int i=head[s]; ~i; i=E[i].nxt)
{
int v=E[i].to;
if(d[v]==d[s]+1&&E[i].cap>0)
{
int d=dfs(v,t,min(f,E[i].cap));
if(d>0)
{
E[i].cap-=d;
E[i^1].cap+=d;
f-=d;
ret+=d;
if(!f)
break;
}
}
}
if(!ret)
d[s]=-1;
return ret;
}
int dinic(int s,int t)
{
int ret=0;
while (bfs(s,t))
ret+=dfs(s,t,INF);
return ret;
}
int main(void)
{
int n,m,b,k,i;
while (~scanf("%d%d",&n,&m))
{
init();
int S=0,T=n+m+1;
for (i=1; i<=n; ++i)
{
scanf("%d",&k);
add(S,i,1);
while (k--)
{
scanf("%d",&b);
add(i,n+b,1);
stall.push_back(b);
}
}
sort(stall.begin(),stall.end());
stall.erase(unique(stall.begin(),stall.end()),stall.end());
int R=stall.size();
for (i=0; i<R; ++i)
add(n+stall[i],T,1);
printf("%d\n",dinic(S,T));
}
return 0;
}