USACO Section 1.3 Calf Flac - 简单的字符串处理

    继续切...这题就是求最长的回文串(忽略奇怪符号..只保留字母,并且不考虑大小写)...枚举中间点..以奇数和偶数的两种情况来搜就可以了...恶心的是有多行..并且多行之间也可以是回文...那就在读的时候将他们都存在一个字符串里..但是' \n ' 也要跟着一起存～～

Program:

/* 
ID: zzyzzy12 
LANG: C++ 
TASK: calfflac
*/  
#include<iostream>  
#include<stdio.h>  
#include<string.h>  
#include<math.h>  
#include<algorithm>  
using namespace std;  
int ans,i,p,h1,h2,len,j,anslen;
char s[20010],str[20010];
int turn(char c)
{
     if (c>='A' && c<='Z') return c-'A';
     if (c>='a' && c<='z') return c-'a';
     return -1;   
}
int main()  
{  
     freopen("calfflac.in","r",stdin);  
     freopen("calfflac.out","w",stdout);  
     len=0;
     while (gets(str))
     {
          p=strlen(str); 
          for (i=0;i<p;i++) s[i+len]=str[i];
          len+=p;  s[len]='\n'; len++;
     }
     ans=0;
     for (i=0;i<len;i++)
     {
           p=1;
           h1=h2=i;
           while (h1>=0 && turn(s[h1])==-1) h1--;
           while (h2<len && turn(s[h2])==-1) h2++;  
           while (h1>=0 && h1<len && turn(s[h1])==turn(s[h2]))
           {
                 if (p>ans)
                 {
                      ans=p;
                      for (j=h1;j<=h2;j++) str[j-h1]=s[j];
                      anslen=h2-h1+1;        
                 } 
                 h1--;
                 while (h1>=0 && turn(s[h1])==-1) h1--;
                 h2++;           
                 while (h2<len && turn(s[h2])==-1) h2++;  
                 p+=2;  
           } 
           p=2;
           h1=i; h2=i+1;
           while (h1>=0 && turn(s[h1])==-1) h1--;
           while (h2<len && turn(s[h2])==-1) h2++;  
           while (h1>=0 && h1<len && turn(s[h1])==turn(s[h2]))
           {
                 if (p>ans)
                 {
                      ans=p;
                      for (j=h1;j<=h2;j++) str[j-h1]=s[j];
                      anslen=h2-h1+1;          
                 } 
                 h1--;
                 while (h1>=0 && turn(s[h1])==-1) h1--;
                 h2++;           
                 while (h2<len && turn(s[h2])==-1) h2++;  
                 p+=2;  
           } 
     }
     printf("%d\n",ans); 
     for (i=0;i<anslen;i++) 
     {
           if (str[i]!='\n')  printf("%c",str[i]); 
           else printf("\n");
     }
     printf("\n");
     return 0;    
}