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

    B. BerSU Ball
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard 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.

    Output

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

    Sample test(s)
    Input
    4
    1 4 6 2
    5
    5 1 5 7 9
    
    Output
    3
    
    Input
    4
    1 2 3 4
    4
    10 11 12 13
    
    Output
    0
    
    Input
    5
    1 1 1 1 1
    3
    1 2 3
    
    Output
    2
    

    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
    ", hungry());
    	}
    	return 0;
    }


    版权声明:本文博主原创文章。博客,未经同意不得转载。

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  • 原文地址:https://www.cnblogs.com/blfshiye/p/4938404.html
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