POJ 1469 COURSES(二部图匹配)

                                                                 COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21527		Accepted: 8460

Description

Consider a group of N students and P courses. Each student visits zero, one or more than one courses. Your task is to determine whether it is possible to form a committee of exactly P students that satisfies simultaneously the conditions:

• every student in the committee represents a different course (a student can represent a course if he/she visits that course)
• each course has a representative in the committee

Input

Your
program should read sets of data from the std input. The first line of
the input contains the number of the data sets. Each data set is
presented in the following format:

P N

Count1 Student_{1 1} Student_{1 2} ... Student_{1 Count1}

Count2 Student_{2 1} Student_{2 2} ... Student_{2 Count2}

...

CountP Student_{P 1} Student_{P 2} ... Student_{P CountP}

The first line in each data set contains two positive integers
separated by one blank: P (1 <= P <= 100) - the number of courses
and N (1 <= N <= 300) - the number of students. The next P lines
describe in sequence of the courses �from course 1 to course P, each
line describing a course. The description of course i is a line that
starts with an integer Count i (0 <= Count i <= N) representing
the number of students visiting course i. Next, after a blank, you抣l
find the Count i students, visiting the course, each two consecutive
separated by one blank. Students are numbered with the positive integers
from 1 to N.

There are no blank lines between consecutive sets of data. Input data are correct.

Output

The
result of the program is on the standard output. For each input data set
the program prints on a single line "YES" if it is possible to form a
committee and "NO" otherwise. There should not be any leading blanks at
the start of the line.

Sample Input

2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1

Sample Output

YES
NO
【分析】最大匹配模板题。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#define inf 0x7fffffff
#define met(a,b) memset(a,b,sizeof a)
typedef long long ll;
using namespace std;
const int N = ;
const int M = ;
int n,m,cnt;
int x[M],y[M];
int t[M],head[N];
struct man
{
    int to,next;
}g[M];
bool dfs(int u) {
    for(int i=head[u];i!=-;i=g[i].next) {
        int v=g[i].to;
        if(!t[v]) {
            t[v]=;
            if(!y[v]||dfs(y[v])) {
                x[u]=v;
                y[v]=u;
                return true;
            }
        }
    }
    return false;
}
void MaxMatch() {
    int ans=;
    for(int i=; i<=m; i++) {
        if(!x[i]) {
            met(t,);
            if(dfs(i))ans++;
        }
    }
    if(ans==m)printf("YES\n");
    else printf("NO\n");
}
void add(int u,int v) {
    g[cnt].to=v;g[cnt].next=head[u];head[u]=cnt++;
}
int main() {
    int T;
    int k,number;
    bool flag;
    scanf("%d",&T);
    for(int k=; k<=T; k++) {
        met(g,);
        met(x,);
        met(y,);
        met(head,-);cnt=;
        scanf("%d%d",&m,&n);
        for(int i=; i<=m; i++) {
            int t,u;
            scanf("%d",&t);
            while(t--) {
                scanf("%d",&u);
                add(i,u);
            }
        }
        MaxMatch();
    }
    return ;
}