【水题】USACO Broken Necklace

进入USACO要注册才可看题: http://train.usaco.org/usacogate

题目：【已翻译】http://www.wzoi.org/usaco/11%5C202.asp

SAMPLE INPUT (file beads.in)
29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

SAMPLE OUTPUT (file beads.out)
11

又一水题……YY的算法//无法用语言表达啊

……感觉飞一般的繁琐！
虽然1次A……但是写代码时感觉很是不爽

/*
ID: 1006100071
PROG: beads
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
#include <set>
//#include <map>
#include <queue>
#include <utility>
#include <iomanip>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <ctype.h>
using namespace std;

int main()
{
	freopen ("beads.in", "r", stdin);
	freopen ("beads.out", "w", stdout);
	int i, len, maxs, start, change;
	bool hash[30] = {0};
	string s;
	cin >> len >> s;
	s = s + s;    //实现首尾相接
	maxs = 1, change = start = 0;
	len <<= 1;    //串长加倍
	if (s[0] != 'w')
		hash[s[0]-'a'] = true, change = 1;
	for (i = 1; i < len; i++)
	{
		//cout << i << ' ' << change << ' ' << s[i]-'a' << endl;
		if (s[i] != 'w' && !hash[s[i]-'a'])
			memset (hash, false, sizeof(hash)), hash[s[i]-'a'] = true, change++;
		if (change > 2 || i == len-1)
		{
			//cout << i - start << "-----------\n";
			if (maxs < i - start)
				maxs = i - start;
			change = 0;
			memset (hash, false, sizeof(hash));
			i = start;
			start++;
			if (s[start] != 'w')
				hash[s[start]-'a'] = true, change = 1;
		}
	}
	if (maxs > len / 2)   //最多取所给原来的串长
		maxs = len / 2;
	cout << maxs << endl;
	return 0;
}