Theme Section HDU - 4763 （kmp）

最新推荐文章于 2019-05-29 00:11:18 发布

原创最新推荐文章于 2019-05-29 00:11:18 发布 · 247 阅读

0 ·

CC 4.0 BY-SA版权

----kmp 专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一种算法，用于解决音乐节中寻找每首歌曲最大主题段的问题。该算法通过预处理找出字符串中的最长重复前缀，并验证是否能在歌曲中找到匹配的三段重复主题。适用于处理大型数据集。

Theme Section

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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

0
0
1
1
2

题意是保证前缀后缀相同的情况下，求这个前缀或者后缀长度最多能多长并保证前缀后缀中间还有一个这样的与前缀后缀相同的字符串。

参考 http://blog.youkuaiyun.com/chilumanxi/article/details/50858269

看别人的用这样的方法。也是可以用这种复杂度过的。

#include<algorithm>
#include<vector>
#include<cstring>
#include<string>
#include<iomanip>
#include<cstdio>
#include<stack>
#include<iostream>
#include<map>
#include<queue>
#include<cmath>
#include<strstream>
using namespace std;
#define sf scanf
#define pf printf
#define mem(a,b) memset(a,b,sizeof(a));
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define MP make_pair
#define N 1000050
#define M 200020
#define mod 998244353
#define ULL unsigned long long
#define LL long long
#define inf 0x3f3f3f3f
//freopen("in.txt","r",stdin);

int t;
char s[N];
int nxt[N];
int n;
bool Find(){
    int j=0;
    for(int i=t;i<n-t;++i){
        while(j&&s[j]!=s[i])j=nxt[j];
        if(s[i]==s[j])++j;
        if(j==t)return true;
    }
    return false;
}
int main(){
    freopen("in.txt","r",stdin);
    int T;sf("%d",&T);
    while(T--){
        mem(nxt,0);
        sf("%s",s);n=strlen(s);
        nxt[0]=0;nxt[1]=0;
        for(int i=1;i<n;++i){
            int j=nxt[i];
            while(j&&s[i]!=s[j])j=nxt[j];
            nxt[i+1]=(s[i]==s[j]?j+1:0);
        }
         t=nxt[n];
         int ans=0;
         while(t){
             if(n>=3*t&&Find()){
                 ans=t;break;
             }
             t=nxt[t];
         }
         pf("%d\n",ans);
    }
}