poj 1469 COURSES（判斷是否是存在不匹配的點（對於右邊的點集來說））

COURSES

點擊打開題目鏈接

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17858		Accepted: 7038

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

題目大意：有P門課和N個學生，第i門課有counti個學生學習；每個學習i課的學生均可代表這個課；

問能否找出學生，每個學生只代表一門課，並且，沒有課時沒有學生代表的；

題目大概是求，對於一個二分圖的匹配，對於右邊的集合（或者左邊的集合）中的點，全是匹配點，不存在不匹配的點；

用匈牙利算法找出最大匹配後，便利右邊集合中的點（或者左邊集合中的點），看是否存在不匹配的點；

注意：此題如果用右集合中的點來考慮的話，注意在輸入時應將順序反一下。

代碼：

#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
#define N 310
using namespace std;
int w[N][N];
int l[N];
int visit[N],v1,v2;
bool dfs(int x)
{
    int i;
    for(i=1;i<=v2;i++)
    {
        if(!visit[i]&&w[x][i])
        {
            visit[i]=1;
            if(l[i]==0||dfs(l[i]))
            {
                l[i]=x;
                return true;
            }
        }
    }
    return false;
}
int nmatch()
{
    int i;
    for(i=1;i<=v1;i++)
    {
        memset(visit,0,sizeof(visit));
        dfs(i);
    }
    for(i=1;i<=v2;i++)
    {
        if(!l[i])//遍歷點看是否存在不匹配的點
            return 0;
          //  printf("l[i] %d\n",l[i]);
    }
    return 1;
}
int main()
{
    int t,p,q,n,i,j,d;
    int ans;
    scanf("%d",&t);
    {
        while(t--)
        {
            scanf("%d%d",&p,&n);
            v1=n,v2=p;//v1，v2的大小也要換一下
            memset(w,0,sizeof(w));
            memset(l,0,sizeof(l));
            for(i=1;i<=p;i++)
            {
                scanf("%d",&q);
                for(j=0;j<q;j++)
                {
                    scanf("%d",&d);
                    w[d][i]=1;///順序反一下（注意到這應該就沒問題了）
                }
            }
            ans=nmatch();
            printf("%s\n",ans==1?"YES":"NO");
        }
    }
    return 0;
}