Codeforces Round #277.5 (Div. 2)B——BerSU Ball

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#二分匹配 #codeforces

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本文介绍了一种通过二分匹配算法解决舞会中男女舞伴技能匹配问题的方法。该算法考虑了男女双方舞蹈技能的差异，并确保每一对舞伴间的技能差距不超过一。通过具体的示例和代码实现展示了如何最大化匹配的舞伴对数。

The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.

We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.

For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequence a₁, a₂, ..., a_n (1 ≤ a_i ≤ 100), where a_i is the i-th boy's dancing skill.

Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b₁, b₂, ..., b_m (1 ≤ b_j ≤ 100), where b_j is the j-th girl's dancing skill.

Output

Print a single number — the required maximum possible number of pairs.

Sample test(s)

Input

Output

Input

4
1 2 3 4
4
10 11 12 13

Output

Input

Output

二分匹配模板题

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;


const int N = 110;

int mark[N];
bool vis[N];
int head[N];
int tot;
int n, m;
int b[N];
int g[N];

struct node
{
	int next;
	int to;
}edge[N * N];

void addedge(int from, int to)
{
	edge[tot].to = to;
	edge[tot].next = head[from];
	head[from] = tot++;
}

bool dfs(int u)
{
	for (int i = head[u]; ~i; i = edge[i].next)
	{
		int v = edge[i].to;
		if (!vis[v])
		{
			vis[v] = 1;
			if (mark[v] == -1 || dfs(mark[v]))
			{
				mark[v] = u;
				return 1;
			}
		}
	}
	return 0;
}

int hungry()
{
	memset(mark, -1, sizeof(mark));
	int ans = 0;
	for (int i = 1; i <= n; ++i)
	{
		memset(vis, 0, sizeof(vis));
		if (dfs(i))
		{
			ans++;
		}
	}
	return ans;
}

int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &b[i]);
		}
		scanf("%d", &m);
		for (int i = 1; i <= m; ++i)
		{
			scanf("%d", &g[i]);
		}
		memset (head, -1, sizeof(head));
		tot = 0;
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= m; ++j)
			{
				if(abs(b[i] - g[j]) <= 1)
				{
					addedge(i, j);
				}
			}
		}
		printf("%d\n", hungry());
	}
	return 0;
}