Time Limit: 1000MS Memory Limit: 10000K

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 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


大意是有N个学生P门课,每个学生可以选某些课,问是否有办法让每门课都至少被一个学生选。学生和课构成了一个二分图,如果最大匹配包含了全部P门课则可以达到目标,即转化为求二分图最大匹配。
AC代码:
//邻接矩阵
#include <cstdio>
#include <cstring>
using namespace std;
int p, n;
const int maxp = 110;
const int maxn = 310;
bool g[maxp][maxn];
int linker[maxn];
bool vis[maxn];
void input()
{
	scanf("%d%d", &p, &n);
	int t, v;
	memset(g, 0, sizeof(g));
	for (int i = 1; i <= p; i++)
	{
		scanf("%d", &t);
		while (t--)
		{
			scanf("%d", &v);
			g[i][v] = 1;
		}
	}
}
bool dfs(int u)
{
	for (int v = 1; v <= n; v++)
		if (g[u][v] && !vis[v])
		{
			vis[v] = 1;
			if (linker[v] == -1 || dfs(linker[v]))
			{
				linker[v] = u;
				return 1;
			}
		}
	return 0;
}
int hungary()
{
	int res = 0;
	memset(linker, -1, sizeof(linker));
	for (int u = 1; u <= p; u++)
	{
		memset(vis, 0, sizeof(vis));
		if (dfs(u))
			res++;
	}
	return res;
}
void solve()
{
	if (hungary() == p)
		printf("YES\n");
	else
		printf("NO\n");
}
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		input();
		solve();
	}
	return 0;
}
//邻接表
#include <cstdio>
#include <cstring>
using namespace std;
int p, n;
const int maxn = 310;
struct edge
{
	int v, next;
}e[30010];
int linker[maxn];
bool vis[maxn];
int k, head[30010];
void adde(int u, int v)
{
	e[k].v = v;
	e[k].next = head[u];
	head[u] = k++;
}
void input()
{
	scanf("%d%d", &p, &n);
	int t, v;
	k = 1;
	memset(head, -1, sizeof(head));
	for (int i = 1; i <= p; i++)
	{
		scanf("%d", &t);
		while (t--)
		{
			scanf("%d", &v);
			adde(i, v);
		}
	}
}
bool dfs(int u)
{
	for (int i = head[u]; i != -1; i = e[i].next)
	{
		int v = e[i].v;
		if (!vis[v])
		{
			vis[v] = 1;
			if (linker[v] == -1 || dfs(linker[v]))
			{
				linker[v] = u;
				return 1;
			}
		}
	}
	return 0;
}
int hungary()
{
	int res = 0;
	memset(linker, -1, sizeof(linker));
	for (int u = 1; u <= p; u++)
	{
		memset(vis, 0, sizeof(vis));
		if (dfs(u))
			res++;
	}
	return res;
}
void solve()
{
	if (hungary() == p)
		printf("YES\n");
	else
		printf("NO\n");
}
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		input();
		solve();
	}
	return 0;
}

 

分类: ACM/ICPC

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