二分图最大匹配——POJ 1469

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COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17976		Accepted: 7086

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

题意：有P门课和N个学生，一个学生可以选0~p门课，问是否有一个集合包含P个学生，使得该集合里每个学生分别对应P门课，即每门可都有学生对应

思路：二分图求最大匹配模板，就是用个net[i][j]数组，i表示第i门课程，j表示学生ID。这里用DFS和BFS

DFS

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
#define ms(x,y) memset(x,y,sizeof(x))
const int MAXN=300+10;
const int INF=1<<30;
using namespace std;
int link[MAXN];
int vis[MAXN];
int net[MAXN][MAXN];
int left,right;
int c,n;

int dfs(int u)
{
	for(int v=1; v<=n; v++){
		if(!vis[v] && net[u][v]){
			vis[v] = 1;
			if(link[v] == -1 || dfs(link[v])){
				link[v] = u;
				return 1;
			}
		}
	}
	return 0;
}

int MaxMatch()
{
	int num=0;
	ms(link, -1);
	for(int i=1; i<=c; i++){
		ms(vis, 0);
		if(dfs(i)) num++;
	}
	return num;
}

int main()
{
	//freopen("in.txt","r",stdin);
	int T;
	scanf("%d", &T);
	while(T--)
	{
		ms(net,0);
		scanf("%d%d", &c,&n);
		if(n < c){
			printf("NO\n");
			continue;
		}
		for(int i=1; i<=c; i++){
			int t,m;
			scanf("%d", &t);
			for(int j=0; j<t; j++){
				scanf("%d", &m);
				net[i][m] = 1;
			}
		}
		int ans = MaxMatch();
		if(ans == c) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}

BFS

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
#define ms(x,y) memset(x,y,sizeof(x))
const int MAXN=300+10;
const int INF=1<<30;
using namespace std;
int link_l[MAXN];
int link_r[MAXN];
int vis[MAXN];
int pre[MAXN];
int net[MAXN][MAXN];
int c,n;

int MaxMatch()
{
	ms(link_l, -1);
	ms(link_r, -1);
	int res = 0;
	for(int i=1; i<=c; i++){//以左边(即课程)为起点寻找一条增广路
		ms(vis, 0);//清空标记
		ms(pre, 0);//清空前驱
		queue<int>q;
		q.push(i);
		int ok=0;
		while(!q.empty())
		{
			int u = q.front();
			q.pop();
			for(int v=1; v<=n; v++){
				if(!vis[v] && net[u][v]){//在右边(即学生)寻找
					vis[v] = 1;
					if(link_l[v] == -1){//该点还没有匹配
						ok = 1;
						int l = u, r = v;
						while(l)//利用前驱反转路径
						{
							int tmp = link_r[l];
							link_l[r] = l;
							link_r[l] = r;
							l = pre[l];
							r = tmp;
						}
						break;//已经找到增广路，退出
					}
					else{//该点已经匹配
						pre[link_l[v]] = u;//记录前驱
						q.push(link_l[v]);//继续寻找
					}
				}
			}
			if(ok) break;//已经找到增广路，退出
		}
		if(ok) res++;//有增广路，总值加1
	}
	return res;
}

int main()
{
	//freopen("in.txt","r",stdin);
	int T;
	scanf("%d", &T);
	while(T--)
	{
		ms(net,0);
		scanf("%d%d", &c,&n);
		if(n < c){
			printf("NO\n");
			continue;
		}
		for(int i=1; i<=c; i++){
			int t,m;
			scanf("%d", &t);
			for(int j=0; j<t; j++){
				scanf("%d", &m);
				net[i][m] = 1;
			}
		}
		int ans = MaxMatch();
		if(ans == c) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}