poj1469 二分图的完备匹配

COURSES

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 18398 Accepted: 7249

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

23 33 1 2 32 1 21 13 32 1 32 1 31 1

Sample Output

YESNO

题意：P个人，N门课程，要求P个人每人选一门互不相同的课。

这是很裸的二分图的完备匹配，所谓完备匹配是指匹配数=min(|X|,|Y|),|X|表示X部的点数，|Y|表示Y部的点数.

代码：

#include<cstdio>#include<iostream>#include<cstring>#define Maxn 310using namespace std;int adj[Maxn][Maxn];int match[Maxn];int vis[Maxn];int x,y;int deep;bool dfs(int u){    for(int v=1;v<=y;v++){        if(adj[u][v]&&vis[v]!=deep){            vis[v]=deep;            if(match[v]==-1||dfs(match[v])){                match[v]=u;                return true;            }        }    }    return false;}int hungary(){    memset(match,-1,sizeof match);    memset(vis,-1,sizeof vis);    int ans=0;    for(int i=1;i<=x;i++){        deep=i;        if(dfs(i)) ans++;    }    return ans;}int main(){    int t,k,a;    cin>>t;    while(t--){        scanf("%d%d",&x,&y);        memset(adj,0,sizeof adj);        for(int i=1;i<=x;i++){            scanf("%d",&k);            for(int j=0;j<k;j++){                scanf("%d",&a);                adj[i][a]=1;            }        }        if(hungary()==x) puts("YES");        else puts("NO");    }return 0;}