USACO Broken Necklace

去掉题目的背景：就是一个环形的串中，寻找一个最长的子串，该串由前后由俩部分组成，连续的b串和连续的r串，当然，一种颜色也可以；w可转变成任意颜色；

我的思路：比较简单的思路，但是是一个复杂度O（n^2）的算法；

先求出以每一个位置开始的最长串的长度，再找出可衔接的最长串即可；

USACO上面有更快的O（n)算法。

Holland's Frank Takes has a potentially easier solution: 

/* This solution simply changes the string s into ss, then for every starting
// symbol it checks if it can make a sequence simply by repeatedly checking 
// if a sequence can be found that is longer than the current maximum one.
*/

#include <iostream>
#include <fstream>
using namespace std;

int main() {
  fstream input, output;
  string inputFilename = "beads.in", outputFilename = "beads.out";  
  input.open(inputFilename.c_str(), ios::in);
  output.open(outputFilename.c_str(), ios::out);
  
  int n, max=0, current, state, i, j;
  string s;
  char c;
  
  input >> n >> s;
  s = s+s;
  for(i=0; i<n; i++) {
    c = (char) s[i];
    if(c == 'w')
      state = 0;
    else
      state = 1;
    j = i;
    current = 0;
    while(state <= 2) { 
      // dont go further in second string than starting position in first string
      while(j<n+i && (s[j] == c || s[j] == 'w')) { 
        current++;
        j++;
      } // while
      state++;
      c = s[j];
    } // while
    if(current > max)
      max = current;
  } // for
    
  output << max << endl;
  return 0;
} // main

/*
ID: nanke691
LANG: C++
TASK: beads
*/
#include<iostream>
#include <fstream>
#include<string.h>
using namespace std;
char s1[700];
int dp[355];
int main()
{
	int n;
    FILE *fin  = fopen ("beads.in", "r");
    FILE *fout = fopen ("beads.out", "w");
	//while(scanf("%d",&n)==1)
	//{
	    fscanf(fin,"%d",&n);
		fscanf(fin,"%s",s1);
		for(int i=0;i<n;i++)
			s1[i+n]=s1[i];
		s1[2*n]='\0';
		//puts(s1);
		int flag=0;
		memset(dp,0,sizeof(dp));
		for(int i=0;i<n;i++)
		{	
			if(s1[i]=='w')
				flag=1;
			else flag=0;
			char temp=s1[i];
			for(int j=i+1;j<2*n;j++)
			{
				if(flag&&s1[j]!='w')
				{
					flag=0;
					temp=s1[j];
				}
				if(temp==s1[j]||s1[j]=='w')
					dp[i]++;
				else break;
			}
			dp[i]++;
		}
		//for(int i=0;i<n;i++)
		//	printf("%d ",dp[i]);
		//cout<<endl;
		int max1=-1;
		for(int i=0;i<n;i++)
		{
			int t=dp[i],sum;
			if(t>n) 
				sum=dp[i]+dp[t-n];
			else sum=dp[i]+dp[t+i];
			if(sum>max1)
				max1=sum;
			if(max1>=n)
			{
				max1=n;
				break;
			}
		}
		fprintf(fout,"%d\n",max1);
	//}
	return 0;
}