USACO section 1.3.3

1. 输入带回车符，空格，标点符号的字符串：

while (fin >> temp)    {        str += temp;        str += " ";    }unsigned int len = str.size();str.resize(len - 1);

2. 判断是不是回文

3. 以下是代码：

/*ID: dollar41PROG: calfflacLANG: C++*/#include<iostream>#include<fstream>#include<string.h>#define maxn 20010using namespace std;char str1[maxn],str2[maxn*2];int n,l1,mid,len;int nxt[maxn*2];void Manacher (){    int mx=0,id;    len=-1;    memset(nxt,0,sizeof(nxt));    for(int i=1; i<l1; i++)    {        if(mx>i)            nxt[i]=min(nxt[2*id-i],mx-i);        else nxt[i]=1;        for(; str2[i-nxt[i]]==str2[i+nxt[i]]; nxt[i]++);        if(mx<i+nxt[i])        {            mx=i+nxt[i];            id=i;        }        if(len<nxt[i]-1)        {            len=nxt[i]-1;            mid=i;        }    }}void pre(){    str2[0]='$';    int i=0,t=0,k=1;    while(str1[i]!='\0')    {        if(t)        {            int flag=0;            while(1)            {                if((str1[i]>='A'&&str1[i]<='Z')|| (str1[i]>='a'&&str1[i]<='z'))                    break;                if(str1[i]=='\0')                {                    flag=1;                    break;                }                i++;            }            if(flag)                break;            str2[k++]=str1[i++];        }        else str2[k++]='#';        if(str2[k-1]>='A'&&str2[k-1]<='Z')            str2[k-1]+=32;        t^=1;    }    str2[k++]='#';    str2[k]='\0';    l1=k;}void  print(int s){    int count=0,flag=0;    for(int i=s; str1[i]!='\0'; i++)    {        if((str1[i]>='A'&&str1[i]<='Z')|| (str1[i]>='a'&&str1[i]<='z'))            count++;        printf("%c",str1[i]);        if(count==len)            break;    }    printf("\n");}int main(){    freopen("calfflac.in","r",stdin);    freopen("calfflac.out","w",stdout);    char str[maxn];    while(gets(str))    {        strcat(str1,"\n");        strcat(str1,str);    }    pre();    //printf("%s\n",str2);    Manacher();    printf("%d\n",len);    int count=0,s=(mid-len)/2+1;    for(int i=0; str1[i]!='\0'; i++)    {        if((str1[i]>='A'&&str1[i]<='Z')|| (str1[i]>='a'&&str1[i]<='z'))            count++;        if(count==s)        {            print(i);            break;        }    }    return 0;}

4. 官方代码：

To make the programming easier, we keep two copies of the text: the original, and one with the punctuation stripped out. We find the biggest palindrome in the latter copy, and then figure out which part of the original it corresponds to.

To find the biggest palindrome in the alphabetic text, we look, for each letter in the text, at the biggest palindrome centered on that letter (in the case of an odd length palindrome) or just to the right of that letter (in the case of an even length palindrome).

There are 20,000 letters, and we are assured that no palindrome is more than 2,000 letters long, and we search for both even and odd palindromes, for a total of about 20,000*2,000*2 = 80,000,000 operations, which is perfectly reasonable within the time limit.

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <assert.h>#include <ctype.h>char fulltext[21000];char text[21000];char *pal;int pallen;voidfindpal(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+1;          bkwd >= text && fwd < etext && *fwd == *bkwd; fwd++, bkwd--);bkwd++;len = fwd - bkwd;if(len > pallen) {    pal = bkwd;    pallen = len;}    }}voidmain(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=0, p=fulltext; *p; p++)if(isalpha(*p))    if(i++ == n)break;    assert(*p != '\0');    /* print out the next pallen characters */    for(i=0; i<pallen && *p; p++) {fputc(*p, fout);if(isalpha(*p))    i++;    }    fprintf(fout, "\n");    exit(0);}