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 = 0; i + 2 < len; i++)
    {
        if(str[i] == str[i+2])
        {
            if(len % 2)
                ans = 1;
            else
                ans = 2;
        }
    }
    printf("%d
", ans);
    return 0;
}