二分图最大匹配加邻接表 - Courses - HDU - 1083

Courses

HDU - 1083

Problem 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

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

P N
Count1 Student1 1 Student1 2 … Student1 Count1
Count2 Student2 1 Student2 2 … Student2 Count2
…
CountP StudentP 1 StudentP 2 … StudentP 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’ll 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.

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.

An example of program input and output:

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

将课程和能够代表它的学生链接

struct Edge
{
    int v,net;
}edge[31010];
void add(int from,int to)
{
    edge[++num].v=to;
    edge[num].net=head[from];
    head[from]=num;
}

对于每门课程依次进行查询 如 A 会在自己的代表中选取一名 a 为其所用 但 a 是 B 的唯一学生 这时就需要反馈到 A 并查询 A 是否能够找到自己其他的代表 若找到了 b 则将 b 设为 A 的代表 并将 a 设为 B 的代表 以此类推就是二分图的最大匹配

邻接表代码

#include<stdio.h>
#include<string.h>
int num,match[333],book[333],head[31010];
struct Edge
{
    int v,net;
}edge[31010];
void add(int from,int to)
{
    edge[++num].v=to;
    edge[num].net=head[from];
    head[from]=num;
}
int dfs(int u)
{
    for(int i=head[u];i;i=edge[i].net)
    {
        int l=edge[i].v;
        if(!book[l])
        {
            book[l]=1;
            if(!match[l]||dfs(match[l]))
            {
                match[l]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int t,p,n,m,i,j,a,s;
    scanf("%d",&t);
    while(t--)
    {
        num=s=0;
        memset(head,0,sizeof(head));
        memset(match,0,sizeof(match));
        scanf("%d %d",&p,&n);
        for(i=1; i<=p; i++)
        {
            scanf("%d",&m);
            for(j=0; j<m; j++)
            {
                scanf("%d",&a);
                add(i,a);
            }
        }
        for(i=1; i<=p; i++)
        {
            memset(book,0,sizeof(book));
            if(dfs(i)) s++;
        }
        if(s==p) printf("YES\n");
        else printf("NO\n");
    }
}

常规代码

#include<stdio.h>
#include<string.h>
int n,v[333],book[333],e[110][333];
int dfs(int u)
{
	for(int i=1;i<=n;i++)
	{
		if(book[i]==0&&e[u][i]==1)
		{
			book[i]=1;
			if(v[i]==0||dfs(v[i]))
			{
				v[i]=u;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	int t,p,m,i,j,a,s;
	scanf("%d",&t);
	while(t--)
	{
		s=0;
		scanf("%d %d",&p,&n);
		memset(e,0,sizeof(e));
		for(i=1; i<=p; i++)
		{
			scanf("%d",&m);
			for(j=0; j<m; j++)
			{
				scanf("%d",&a);
				e[i][a]=1;
			}
		}
		memset(v,0,sizeof(v));
		for(i=1; i<=p; i++)
		{
			memset(book,0,sizeof(book));
			if(dfs(i)) s++;
		}
		if(s==p) printf("YES\n");
		else printf("NO\n");
	}
}