HDU4763-Theme Section（KMP+二分）

音乐节主题段落算法

最新推荐文章于 2025-09-12 18:08:49 发布

转载最新推荐文章于 2025-09-12 18:08:49 发布 · 67 阅读

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原文链接：http://www.cnblogs.com/yfceshi/p/6930679.html

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本文介绍了一种算法，用于解决音乐节上歌曲的主题部分长度最大化的问题。通过二分查找结合KMP算法来确定每首歌中主题部分的最大长度。

Theme Section

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1098 Accepted Submission(s): 570

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

题意：输出最长的前缀，前缀要满足在原串中至少匹配3次

思路：二分前缀的结尾点。右边为3分之中的一个原串，用KMP 统计出现次数。就可以

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
const int  maxn = 1000000+10;
int next[maxn],n;
char str[maxn];

void getNext(){
    next[0] = next[1] = 0;
    for(int i = 1,j; i < n; i++){
        j = next[i];
        while(j && str[i] != str[j]) j = next[j];
        if(str[i] == str[j]) next[i+1] = j+1;
        else next[i+1] = 0;
    }
}

bool find(int ed){
    int cnt = 0;
    for(int i = 0,j = 0; i < n; i++){
        while(j && str[i] != str[j]) j = next[j];
        if(str[i]==str[j]) j++;
        if(j==ed){
            ++cnt;
            j = 0;
            if(cnt>=3) return true;
        }
    }
    return false;
}

int binary_search(){
    int L = 1,R = n/3+1,mid;
    while(L <= R){
        mid = (L+R)>>1;
        if(find(mid)){
            L = mid+1;
        }else{
            R = mid-1;
        }
    }
    if(find(L)) return L;
    else if(find(R)) return R;
    else return  L-1;

}

int main(){

    int ncase;
    cin >>  ncase;
    while(ncase--){
        scanf("%s",str);
        n = strlen(str);
        getNext();
        printf("%d\n",binary_search());
    }
    return 0;
}

转载于:https://www.cnblogs.com/yfceshi/p/6930679.html