usaco Calf Flac

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这题考察最基本的回文串处理，采取的算法是枚举回文串的中心点，不过要分回文串长度是奇数和偶数的情况。

/*ID: ljracm1LANG: C++PROB: calfflac*/#include<cstdio>#include<cstring>#include<ctype.h>#define Tusing namespace std;int main(){    #ifdef  T    freopen("calfflac.in","r",stdin);    freopen("calfflac.out","w",stdout);    #endif    char ch;    char str1[20005],str[20005];    int len1=0,len=0,h[20005]; //h[20005]专门用来记录原字符串中非标点字符的实际位置    while(scanf("%c",&ch)!=EOF){         if((ch>='A'&&ch<='Z')||(ch>='a'&&ch<='z')){             h[len1]=len;             str[len++]=ch;             str1[len1++]=tolower(ch);             str[len]='\0';             str1[len]='\0';         }         else{             str[len++]=ch;             str[len]='\0';         }    }    int maxn1 = 0, center1 =0;//maxn1记录最长回文串为奇数的情况, center1为中心节点的起点    for(int i=0; i<len1; i++){//枚举中心节点的情况        int l=0;//从起始长度为0算起        while(str1[l+i]==str1[i-l]&&l+i<len1&&i-l>=0)l++;        if(maxn1<l){            maxn1 = l;            center1 = i;        }    }    int maxn2 = 0, center2 = 0;    for(int i=0; i<len1; i++){        int l=0;        while(str1[l+1+i]==str1[i-l]&&l+i<len1&&i-l>=0)l++;        if(maxn2<l){             maxn2 = l;             center2 = i;        }    }    //前面maxn1与maxn2均是一半长的情况,将它变回实际长度进行比较    maxn1 = maxn1*2-1;    maxn2 *= 2;    int maxn,center;    if(maxn1>maxn2){         maxn = maxn1;  center = center1;    }    else{         maxn = maxn2; center = center2;    }    printf("%d\n",maxn);    for(int i=h[center-(maxn+1)/2+1]; i<=h[center-(maxn+1)/2+maxn]; i++)         printf("%c",str[i]);    printf("\n");    return 0;}

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