USACO2012 Broken necklace /// DP oj10103

题目大意：

项链最长的纯色连续段，“w”即white白色珠子，可任意涂为红或蓝，“r” “b”即红 蓝。

Input

Line 1:  N, the number of beads

Line 2:  a string of N characters, each of which is r, b, or w

Output

A single line containing the maximum of number of beads that can be collected from the supplied necklace.

Sample Input

29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

Sample Output

11

DP算法

*先复制相同的一段 拼接在一起 以模拟项链的环状

#include <bits/stdc++.h>
using namespace std;
typedef struct node
{
    int lr,lb,rr,rb;            ///左右的red blue珠数量
}beans;
int main()
{
    beans bean[800];
    char br[800],brr[400];
    int n; scanf("%d%s",&n,br);
    strcpy(brr,br);
    strcat(br,brr);

    bean[0].lr=bean[0].lb=0;
    for(int i=1;i<=2*n;i++)
    {
        if(br[i-1]=='r')
            bean[i].lr=bean[i-1].lr+1,
            bean[i].lb=0;
        else if(br[i-1]=='b')
            bean[i].lb=bean[i-1].lb+1,
            bean[i].lr=0;
        else
            bean[i].lb=bean[i-1].lb+1,
            bean[i].lr=bean[i-1].lr+1;
    }

    bean[2*n].rr=bean[2*n].rb=0;
    for(int i=2*n-1;i>=0;i--)
    {
        if(br[i]=='r')
            bean[i].rr=bean[i+1].rr+1,
            bean[i].rb=0;
        else if(br[i]=='b')
            bean[i].rb=bean[i+1].rb+1,
            bean[i].rr=0;
        else
            bean[i].rb=bean[i+1].rb+1,
            bean[i].rr=bean[i+1].rr+1;
    }

    int m=0;
    for(int i=0;i<2*n;i++)
        m=max(m,max(bean[i].lr,bean[i].lb)+max(bean[i].rr,bean[i].rb));
    if(m>n) m=n;
    printf("%d
",m);
    return 0;
}