poj 1469 COURSES 题解

COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21515		Accepted: 8455

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

Source

Southeastern Europe 2000

——————————————————————————我是分割线——————————————————————————————

二分图最大匹配。

邻接矩阵map[i][j]表示j号喜欢i号课程，然后对课程进行寻找匹配，匹配成功则标记，之后统计匹配数目即可。

这题丧心病狂卡时，我改了两天，最后发现I/O超时了。。。 取消同步后的流还是略微慢一些，用标准输入输出可以正好卡过。TAT

/*
    Problem:
    OJ:     
    User:   
    Time:   
    Memory:  
    Length:  
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<iomanip>
#include<cassert>
#include<climits>
#include<vector>
#include<list>
#include<map>
#define maxn 1001
#define F(i,j,k) for(int i=j;i<=k;i++)
#define M(a,b) memset(a,b,sizeof(a))
#define FF(i,j,k) for(int i=j;i>=k;i--)
#define inf 0x7fffffff
#define maxm 2016
#define mod 1000000007
//#define LOCAL
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,p;
bool kc[maxn][maxn];
bool vis[maxn];
int pp[maxn];
inline int path(int u)
{
    int a,b,temp;
    F(i,1,n){
        if(kc[u][i]&&!vis[i]){
            vis[i]=true;
            temp=pp[i];
            pp[i]=u;
            if(temp==-1||path(temp)) return 1;
            pp[i]=temp;
        }
    }
    return 0;
}
inline int solve()
{
    int a,b,ans=0;
    M(pp,-1);
    F(i,1,p){
        M(vis,0);
//        ans+=path(i);
if(path(i)) ans++;
        if(ans==p) break;
    }
    return ans;
}
int main()
{
    int t;cin>>t;
    while(t--){
        scanf("%d", &p);scanf("%d", &n);
        M(kc,0);
        int num,cnt;
        F(i,1,p){
            scanf("%d", &num);
            F(j,1,num){
                scanf("%d", &cnt);
                kc[i][cnt]=true;
            }
        }
        if(solve()==p) printf("YES
");
        else printf("NO
");
    }
    return 0;
}