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  • 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

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<algorithm>
     5 #include<queue>
     6 
     7 using namespace std;
     8 
     9 #define N 1100
    10 #define INF 0xfffffff
    11 #define eps 1e-8
    12 
    13 int main()
    14 {
    15     char str[N];
    16 
    17     scanf("%s", str);
    18     int ans, len;
    19     len = ans = strlen(str);
    20 
    21     for(int i = 0; i + 2 < len; i++)
    22     {
    23         if(str[i] == str[i+2])
    24         {
    25             if(len % 2)
    26                 ans = 1;
    27             else
    28                 ans = 2;
    29         }
    30     }
    31     printf("%d
    ", ans);
    32     return 0;
    33 }
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  • 原文地址:https://www.cnblogs.com/Tinamei/p/4797026.html
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