1 second
256 megabytes
standard input
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.
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.
Print a single number — the required maximum possible number of pairs.
4 1 4 6 2 5 5 1 5 7 9
3
4 1 2 3 4 4 10 11 12 13
0
5 1 1 1 1 1 3 1 2 3
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\n", hungry());
}
return 0;
}