String reduction （poj 3401

String reduction

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1360		Accepted: 447

Description

There is a string of characters 'a' and 'b' with the length of no more than 255 characters. You can perform the substring reduction on the initial string in the following way: a substring "a*a" or "b*b" (where * (asterisk) denotes any character) can be reduces to a substring "*".

The task is to achieve a string of minimal possible length after several substring reductions.

Input

The input contains the initial string.

Output

The output contains a single line with the minimal possible length.

Sample Input

aab

Sample Output

3

Source

Northeastern Europe 2001, Western Subregion

题意：在一个串中形如a*a 或b*b可以变换为a或b，问变换之后最短长度。

由于a*a, b*b之类的字符可以匹配任意字符，那么我们只要从前

向后扫描，如果有相隔的两个字符相同，由于这三个字符能够匹配a,b任意字符，那么我们就可以让这种匹配一直进行

下去，从而将字符串不断的reduction ，只要注意字符串长度的奇偶性就可以了，如此我们可以得到一个很简单的方法

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>

 using namespace std;

 #define N 1100
 #define INF 0xfffffff
 #define eps 1e-8

 int main()
 {
     char str[N];

     scanf("%s", str);
     int ans, len;
     len = ans = strlen(str);

     for(int i = ; i +  < len; i++)
     {
         if(str[i] == str[i+])
         {
             if(len % )
                 ans = ;
             else
                 ans = ;
         }
     }
     printf("%d\n", ans);
     return ;
 }