POJ 1274 The Perfect Stall

The Perfect Stall

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15710		Accepted: 7176

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

今天下午大一排位赛的E题，直接套匈牙利算法模板即可
此题测试数据可能有多组，所以要用while(cin)来写（害的我WA三次T^T）
机房只能用VC6.0，好旧的编译器，所以代码写的可能有点丑

#include<iostream>

using namespace std;

bool map[201][201];
bool mask[201];
int nx,ny;
int cx[201],cy[201];

int findpath(int u)
{
    int i;
    for(i=1;i<=ny;i++)
        if(map[u][i]&&!mask[i])
        {
            mask[i]=true;
            if(cy[i]==-1||findpath(cy[i]))
            {
                cy[i]=u;
                cx[u]=i;
                return 1;
            }
        }
    return 0;
}

int maxMarch()
{
    int res=0;
    int i,j;
    for(i=1;i<=nx;i++)
        cx[i]=-1;
    for(i=1;i<=ny;i++)
        cy[i]=-1;
    for(i=1;i<=nx;i++)
    {
        if(cx[i]==-1)
        {
            for(j=1;j<=ny;j++)
                mask[j]=false;
            res+=findpath(i);
        }
    }
    return res;
}

int main()
{
    int i,j;
    while(cin>>nx>>ny)
    {
        for(i=1;i<=nx;i++)
            for(j=1;j<=ny;j++)
                map[i][j]=false;
        for(i=1;i<=nx;i++)
        {
            int temp;
            cin>>temp;
            for(j=1;j<=temp;j++)
            {
                int tmp;
                cin>>tmp;
                map[i][tmp]=true;
            }
        }
        cout<<maxMarch()<<endl;
    }

    return 0;
}