标签：names ams begin -- main asi sub scanf name
http://acm.hdu.edu.cn/showproblem.php?pid=4763
Theme Section
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5815 Accepted Submission(s): 2890
Problem Description
It‘s
time for music! A lot of popular musicians are invited to join us in
the music festival. Each of them will play one of their representative
songs. To make the programs more interesting and challenging, the hosts
are going to add some constraints to the rhythm of the songs, i.e., each
song is required to have a ‘theme section‘. The theme section shall be
played at the beginning, the middle, and the end of each song. More
specifically, given a theme section E, the song will be in the format of
‘EAEBE‘, where section A and section B could have arbitrary number of
notes. Note that there are 26 types of notes, denoted by lower case
letters ‘a‘ - ‘z‘.
To get well prepared for the festival, the
hosts want to know the maximum possible length of the theme section of
each song. Can you help us?
Input
The
integer N in the first line denotes the total number of songs in the
festival. Each of the following N lines consists of one string,
indicating the notes of the i-th (1 <= i <= N) song. The length of
the string will not exceed 10^6.
Output
There
will be N lines in the output, where the i-th line denotes the maximum
possible length of the theme section of the i-th song.
Sample Input
5
xy
abc
aaa
aaaaba
aaxoaaaaa
Sample Output
Source
Recommend
liuyiding
题意：求前中后最长相同串长度，不能有重叠，
我还以为我这种方法会超时，没想到过了。！！
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
#define INF 10000000
using namespace std;
char a[1000009] , b[1000009], str[200009];
int ans = 0 ;
void getnext(char *a , int len , int *next)
{
next[0] = -1 ;
int k = -1 , j = 0 ;
while(j < len)
{
if(k == -1 || a[j] == a[k])
{
k++;
j++;
next[j] = k ;
}
else
{
k = next[k];
}
}
}
int main()
{
int n ;
scanf("%d" , &n);
while(n--)
{
int next[1000009];
int next1[1000009];
scanf("%s" , a);
int len = strlen(a);
getnext(a , len , next);
int q = next[len];
while(q > 0)
{
if(q * 3 > len)
{
q = next[q];
continue ;
}
for(int i = 0 ; i < q ; i++)
{
str[i] = a[i] ;
}
getnext(a , q , next1);
int j = 0 , i = q ;
while(i < len - q && j < q)
{
if(j == -1 || str[j] == a[i])
{
i++;
j++;
}
else
{
j = next[j] ;
}
}
if(j == q)
{
ans = q;
break ;
}
q = next[q];
}
printf("%d\n" , q);
}
return 0 ;
}
kmp（前中后最长相同长度）
标签：names ams begin -- main asi sub scanf name
原文地址：https://www.cnblogs.com/nonames/p/11296313.html
