Codeforces Round #277.5 (Div. 2)B??BerSU Ball_html/css_WEB-ITnose

B. BerSU Ball

time limit per test

memory limit per test

input

output

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,?...,?an (1?≤?ai?≤?100), where ai 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,?...,?bm (1?≤?bj?≤?100), where bj is the j-th girl's dancing skill.

立即学习“前端免费学习笔记（深入）”；

Output

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

仿B站视频帧预览插件

仿B站视频帧预览插件

49

查看详情

Sample test(s)

Input

41 4 6 255 1 5 7 9

Output

Input

41 2 3 4410 11 12 13

Output

Input

51 1 1 1 131 2 3

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;}