Two Heaps（Codeforces Round #205 (Div. 2)）

题目链接：http://codeforces.com/contest/353/problem/B

题意：给出2n2n个两位数，现在要将其分成两个集合，每个集合nn个数，分好集合后每次可以从两个集合分别取出一个数字组成一个四位数，问如果分可以使得组成的数字种类最多

思路：对于数量超过11的数字，先给每个集合一个，对于数量为11的数字尽量等分到两个集合，剩下的数字随便给

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f, maxn = 205;
int n, a[maxn], ans[maxn], num[maxn], vis[maxn];
int main()
{
	while (~scanf("%d", &n))
	{
		memset(num, 0, sizeof(num));
		memset(ans, 0, sizeof(ans));
		memset(vis, 0, sizeof(vis));
		for (int i = 1; i <= 2 * n; i++)
		{
			scanf("%d", &a[i]);
			num[a[i]]++;
		}
		int num0 = 0, num1 = 0, num2 = 0;
		for (int i = 1; i <= 2 * n; i++)
		{
			if (num[a[i]] >= 2)
			{
				if (!vis[a[i]])num0++, ans[i] = 1, vis[a[i]] = 1;
				else if (vis[a[i]] == 1)ans[i] = 2, vis[a[i]] = 2;
			}
			else
			{
				if (num1 > num2)ans[i] = 2, num2++;
				else ans[i] = 1, num1++;
			}
		}
		num1 += num0, num2 += num0;
		printf("%d\n", num1 * num2);
		for (int i = 1; i <= 2 * n; i++)
			if (vis[a[i]] == 2 && !ans[i])
			{
				if (num1 < n)ans[i] = 1, num1++;
				else ans[i] = 2;
			}
		for (int i = 1; i <= 2 * n; i++)printf("%d%c", ans[i], i == 2 * n ? '\n' : ' ');
	}
	return 0;
}