USACO section1.1 Broken Necklace

这道题理解了题目的含义就很简单了；

如果题目改成每次只统计一端，遇到不同时开始统计另一端，并且只有相邻两次统计的（可以转化）为同一种颜色时，才能累加，难度会很高，可能要DP了；

/*
PROG: beads
LANG: C++
*/
# include <cstdio>
# include <cstring>

# define N 350 + 10

int n;
char s[N];

int count(char *s, int k)
{
    int ret = 0;

    int i = k;
    int j = (k-1+n) % n;

    char cur;
    while(ret < n && s[i] == 'w') {++ret; i = (i+1)%n;}
    cur = s[i];
    while (ret < n && (s[i]==cur || s[i]=='w')) {++ret; i = (i+1)%n;}
    while (ret < n && s[j] == 'w') {++ret; j = (j+n-1)%n;}
    cur = s[j];
    while (ret < n && (s[j]==cur || s[j]=='w')) {++ret; j = (j-1+n)%n;}

    return ret;
}

int main()
{
    freopen("beads.in", "r", stdin);
    freopen("beads.out", "w", stdout);

    int max = 0;
    scanf("%d%s", &n, s);
    for (int i = 0; s[i]; ++i)
    {
        int cur = count(s, i);
        if (cur > max) max = cur;
    }
    printf("%d\n", max);

    fclose(stdin);
    fclose(stdout);

    return 0;
}

/**/