poj 1469 COURSES (二分图模板应用 【*模板】 )

COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18454		Accepted: 7275

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

题目解读：p门课，n个学生. 接下来p行，每行代表第i门课每行先输入这门课的学生数，然后在一次输入这些学生的编号。通过匈牙利算法问：能不能保证每门课至少都有一个学生. 算法要点：最大匹配数>=课程数p ？

 匈牙利算法 代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

vector<int>g[310];
int link[310], vis[310];
int p, n;

bool match(int x)
{
	for(int i=0; i<g[x].size(); i++ )
	{
		if(!vis[g[x][i]] )
		{
			vis[g[x][i]] = true;
			if(link[g[x][i]]==-1 || match(link[g[x][i]]) )
			{
				link[g[x][i]] = x;
				return true;
			}
		}
	}
	return false;
}


int hungary()
{
	int tot=0;
	memset(link, 255, sizeof(link));
	for(int i=1; i<=n; i++)
	{
		memset(vis, 0, sizeof(vis));
		if(match(i) )
		{
			tot++;
		}
	}
    return tot;
}
int main()
{
	int t;
	int i, j, k;

	scanf("%d", &t);
	while(t--)
	{
		scanf("%d %d", &p, &n);
		int dd, u;
		for(i=1; i<=n; i++)
            g[i].clear();
		for(i=1; i<=p; i++)
		{
			scanf("%d", &dd);
			while(dd--)
			{
				scanf("%d", &u);
				g[u].push_back(i);
			}
		}
		int ans = hungary();
        if(ans >= p)
            printf("YES\n");
        else
            printf("NO\n");
	}
	return 0;
}

 Hopcroft-Karp 算法：

 Hopcroft-Karp算法相比普通的匈牙利算法更快，所以当两边集合的点比较多时，为了快速完成匹配可以考虑这个算法，即使是有模板，但代码比较长且比较繁琐，容易写错。

敲的时候要特别注意！

H-K算法代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;
int p, n;
vector<int>g[310];
int n1, n2;
int mx[310], my[310];
queue<int>que;

int dx[310], dy[310];
bool vis[310];


bool find(int u)
{
	for(int i=0; i<g[u].size(); i++)
	{
		if(!vis[g[u][i]] && dy[g[u][i]] == dx[u]+1 )
		{
			vis[g[u][i]] = true;
			if(!my[g[u][i]] || find(my[g[u][i]]) )
			{
				mx[u] = g[u][i];
				my[g[u][i]] = u;
				return true;
			}
		}
	}
	return false;
}

int matching()
{
	memset(mx, 0, sizeof(mx));
	memset(my, 0, sizeof(my));
	int ans=0;

	while(true)
	{
		bool flag=false;
		while(!que.empty())
			que.pop();
		memset(dx, 0, sizeof(dx));
		memset(dy, 0, sizeof(dy));
		for(int i=1; i<=n1; i++)
			if(!mx[i] )
				que.push(i);
		while(!que.empty() )
		{
			int u=que.front();
			que.pop();
			for(int i=0; i<g[u].size(); i++ )
			{
				if(!dy[g[u][i]] )
				{
					dy[g[u][i]] = dx[u]+1;
					if(my[g[u][i]])
					{
						dx[my[g[u][i]]] = dy[g[u][i]] + 1;
						que.push(my[g[u][i]] );
					}
					else
						flag=true;
				}
			}
		}
		if(!flag) break;
		memset(vis, false, sizeof(vis));
		for(int i=1; i<=n1; i++)
		{
			if(!mx[i] && find(i) )
				ans++;
		}
	}
	return ans;
}

int main()
{
	int t;
	scanf("%d", &t);
	int dd, u;
	while(t--)
	{
		scanf("%d %d", &p, &n);
		for(int i=1; i<=n; i++)
            g[i].clear();
		for(int i=1; i<=p; i++)
		{
			scanf("%d", &dd);
			while(dd--)
			{
				scanf("%d", &u);
				g[u].push_back(i);
			}
		}
		n1=n; n2=p;
		int ans = matching();
		if(ans >= p )
		  printf("YES\n");
        else
        printf("NO\n");

	}
	return 0;
}