Codeforces Round #277.5 (Div. 2)B—

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

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发布时间：2019-06-20

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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

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 #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                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;}

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