USACO beads

　　题意是给你一串首尾相连的珠子， 珠子有b r w三种颜色， 然后让你统计某两个珠子中间到两边相同珠子的最大数量是多少？？刚开始我选择了暴力求解， 后来看题解后有一种dp的方法， 感觉蛮赞的。。代码如下：

/*
    ID: m1500293
    LANG: C++
    PROG: beads
*/

#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
const int maxn = 2*400 + 10;

int len;
char str[maxn];
int bl[maxn], br[maxn], rl[maxn], rr[maxn];
int main()
{
    freopen("beads.in", "r", stdin);
    freopen("beads.out", "w", stdout);
    while(scanf("%d", &len) == 1)
    {
        scanf("%s", str+1);
        for(int i=len+1; i<=2*len; i++)
            str[i] = str[i-len];
        bl[0] = rl[0] = 0;
        for(int i=1; i<=2*len; i++)
        {
            if(str[i] == 'r') rl[i]=rl[i-1]+1, bl[i]=0;
            if(str[i] == 'b') bl[i]=bl[i-1]+1, rl[i]=0;
            if(str[i] == 'w') bl[i]=bl[i-1]+1, rl[i]=rl[i-1]+1; 
        }
        br[2*len+1] = rr[2*len+1] = 0;
        for(int i=2*len; i>=1; i--)
        {
            if(str[i] == 'r') rr[i] = rr[i+1]+1, br[i]=0;
            if(str[i] == 'b') br[i] = br[i+1]+1, rr[i]=0;
            if(str[i] == 'w') br[i]=br[i+1]+1, rr[i]=rr[i+1]+1;
        } 
        int res = 0;
        for(int i=1; i<2*len; i++)
        {
            res = max(res, max(bl[i], rl[i])+max(br[i+1], rr[i+1]));
        }
        res = min(res, len);
        printf("%d\n", res);
    }    
    return 0;    
}