【USACO】calfflac

关键：以回文中心位置为变量进行遍历

//必须把纯字母先提出来 否则肯能会出现错误 比如： lvlv= 在检查长度4时 lvlv认为不是回文 vlv=认为是回文 但实际上 lvl 出现的要更早一些
//判断回文的方法 可以输入字符串后 左右比较 或者分别正序 逆序 存储 判断是否相等
//我的思路不对 严重超时了 我是以长度为变量进行循环  对于每个长度 每一个不同起始点的序列都需要对 整个序列重新判断一次是否为回文 O(n^3)
//答案中 以中心字母为变量进行循环 只需要对每一个字母做为中心变量遍历一次， 每次的遍历长度小于等于总序列长度  通过不断更新存储最长回文找到最优解 O(n^2)
#include <stdio.h>
#include <string.h>
#include <ctype.h>
typedef struct
{
    char c;
    int n;
}AZ;

int isch(char c)
{
    return ((c <= 'z' && c >= 'a') || (c >= 'A' && c <= 'Z'));
}
int iscalfflac(AZ *c, int l, int r, int * loc) //输入字符 和 左右边界 判断是否为回文
{
    int len = r - l + ;
    int ll, rr;
    int flag = ;
    int is = ; //判断是否进入过相等 防止纯符号
    while(l <= r)
    {
        if(c[l].c == c[r].c || c[l].c == c[r].c - 'A' + 'a' || c[r].c == c[l].c - 'A' + 'a')
        {
            l++; r--; is++;
        }
        else
        {
            flag = ;break;
        }
    }
    return (flag == ) ?  : ;
}
int main()
{
    FILE *in, *out;
    int loc[] = {,,,};
    int i = , j = , length;
    int r, l, rr, ll;
    char c[], c1[], c2[], c3[], c4[];
    AZ ch[];
    in = fopen("calfflac.in", "r");
    out = fopen("calfflac.out", "w");
    while(fscanf(in, "%c", &c[i]) != EOF)
    {
        if(isch(c[i]))
        {
            ch[j].c = c[i];
            ch[j].n = i;
            j++;
        }
        i++;
    }
    length = j; //总输入长度
    for(i = ; i < length; i++)
    {
        if(isupper(ch[i].c))
        {
            ch[i].c = ch[i].c - 'A' + 'a';
        }
        c1[i] = ch[i].c;
        c2[length - i - ] = ch[i].c;
    }

    for(i = length; i > ; i--) //这种方法在大输入时严重超时了
    {
        for(l = ; l <= length - i; l++ )
        {
            r = l + i - ;
            if(iscalfflac(ch, l, r, loc) == )
            {
                fprintf(out, "%d\n", i);
                for(j = ch[l].n ; j <= ch[r].n; j++)
                {
                    fprintf(out, "%c", c[j]);
                }
                fprintf(out, "\n");
                return ;
            }
        }
    }
    fprintf(out, "%d\n", );
    return ;
}

下面是答案的，注意做中心点时要对 奇数 偶数 考虑齐全

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>

char fulltext[];
char text[];

char *pal;
int pallen;

void
findpal(void)
{
    char *p, *fwd, *bkwd, *etext;
    int len;

    etext = text+strlen(text);
    for(p=text; *p; p++) {
    /* try palindrome with *p as center character */
    for(fwd=bkwd=p; bkwd >= text && fwd < etext && *fwd == *bkwd;
                fwd++, bkwd--)
            ;
    bkwd++;
    len = fwd - bkwd;
    if(len > pallen) {
        pal = bkwd;
        pallen = len;
    }

    /* try palindrome with *p as left middle character */
    for(bkwd=p, fwd=p+;
              bkwd >= text && fwd < etext && *fwd == *bkwd; fwd++, bkwd--)
            ;
    bkwd++;
    len = fwd - bkwd;
    if(len > pallen) {
        pal = bkwd;
        pallen = len;
    }
    }
}

void
main(void)
{
    FILE *fin, *fout;
    char *p, *q;
    int c, i, n;

    fin = fopen("calfflac.in", "r");
    fout = fopen("calfflac.out", "w");
    assert(fin != NULL && fout != NULL);

    /* fill fulltext with input, text with just the letters */
    p=fulltext;
    q=text;
    while((c = getc(fin)) != EOF) {
    if(isalpha(c))
        *q++ = tolower(c);
    *p++ = c;
    }
    *p = '\0';
    *q = '\0';

    findpal();

    fprintf(fout, "%d\n", pallen);

    /* find the string we found in the original text
       by finding the nth character */
    n = pal - text;
    for(i=, p=fulltext; *p; p++)
    if(isalpha(*p))
        if(i++ == n)
        break;
    assert(*p != '\0');

    /* print out the next pallen characters */
    for(i=; i<pallen && *p; p++) {
    fputc(*p, fout);
    if(isalpha(*p))
        i++;
    }
    fprintf(fout, "\n");

    exit();
}