Codeforces Round 205 (Div. 2)
E
Antichain
Codeforces
2200
原题
← D→
翻译中…

You have a directed acyclic graph $*G*$, consisting of $*n*$ vertexes, numbered from $0$ to $*n* - 1$. The graph contains $*n*$ edges numbered from $0$ to $*n* - 1$. An edge with number $*i*$ connects vertexes $*i*$ and $(*i* + 1) *mod* *n*$, and it can be directed in either direction (from $*i*$ to $(*i* + 1) *mod* *n*$, or vise versa).

翻译中…

Operation $*x* *mod* *y*$ means taking the remainder after dividing number $*x*$ by number $*y*$.

翻译中…

Let's call two vertexes $*u*$ and $*v*$ in graph $*G*$ comparable if the graph contains a path either from $*u*$ to $*v*$ or from $*v*$ to $*u*$. We'll assume that an antichain is a set of vertexes of graph $*G*$, where any two distinct vertexes are not comparable. The size of an antichain is the number of vertexes in the corresponding set. An antichain is maximum if the graph doesn't have antichains of a larger size.

翻译中…

Your task is to find the size of the maximum antichain in graph $*G*$.

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Input

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The first line contains the sequence of characters $*s*_{0}*s*_{1}... *s*_{*n* - 1}$ $(2 ≤ *n* ≤ 10^{6})$, consisting of numbers zero and one. The length of the line (number $*n*$) corresponds to the number of vertexes and edges in graph $*G*$. If character $*s*_{*i*}$ $(*i* ≥ 0)$ equals $0$, then the edge between vertexes $*i*$ and $(*i* + 1) *mod* *n*$ is directed from the $*i*$-th vertex to the $(*i* + 1) *mod* *n*$-th one, otherwise — to the opposite point.

翻译中…

It is guaranteed that the given graph is acyclic.

翻译中…

Output

翻译中…

Print a single integer — the size of the maximum antichain of graph $*G*$.

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Note

翻译中…

Consider the first test sample. The graph's $*G*$ edges are: $0 → 1$, $1 → 2$, $0 → 2$. We can choose the set of vertexes $[0]$ as the maximum antichain. We cannot choose an antichain of larger size.

翻译中…

样例

Input #1
001
Output #1
1
InputCopy OutputCopy
Input #2
110010
Output #2
3
InputCopy OutputCopy
题目信息
题号
353E
时间限制
1 s
内存限制
256 MB
标签
动态规划
图匹配
贪心

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