Wave hdu-6570

Avin is studying series. A series is called “wave” if the following conditions are satisfied:

1. It contains at least two elements;
2. All elements at odd positions are the same;
3. All elements at even positions are the same;
4. Elements at odd positions are NOT the same as the elements at even positions.
   You are given a series with length n. Avin asks you to find the longest “wave” subseries. A subseries is a subsequence of a series.
   Input
   The first line contains two numbers n, c (1 ≤ n ≤ 100, 000, 1 ≤ c ≤ 100). The second line contains n integers whose range is [1, c], which represents the series. It is guaranteed that there is always a “wave” subseries.
   Output
   Print the length of the longest “wave” subseries.
   Sample Input
   5 3
   1 2 1 3 2
   Sample Output
   4

先把每个数的位置用一个vector存下来，然后从1-c里面找两个数进行双重循环，
x1是vv[i]的size，x2是vv[j]的size，now是当前已经进行到了那个位置
while（vv[i][x1]<now）是找到第一个大于now的位置，如果找到了sum++；

#include<iostream>
#include<vector>
using namespace std;
int main()
{
	int n,c;int x;int res=0;
	vector<int> vv[1234];
	cin>>n>>c;
	for(int i=1;i<=n;i++)
	{
		cin>>x;
		vv[x].push_back(i);
	}
	for(int i=1;i<=c;i++)
	{
		for(int j=1;j<=c;j++)
		{
			if(i==j) continue;
			int x2=0;int x1=0;int now=0;int sum=0;
			for( ; ;)
			{
				while(vv[i][x1]<now&&x1<vv[i].size()) x1++;
				now=vv[i][x1];
				if(x1==vv[i].size()) break;
				sum++;
				while(vv[j][x2]<now&&x2<vv[j].size()) x2++;
				now=vv[j][x2];
				if(x2==vv[j].size()) break;
				sum++;
			}
			res=max(res,sum);
		}
	}
	cout<<res<<endl;
	return 0;
}