Description

A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the piano. It is unfortunate but true that this representation of melodies ignores the notion of musical timing; but, this programming task is about notes and not timings.

Many composers structure their music around a repeating &qout;theme&qout;, which, being a subsequence of an entire melody, is a sequence of integers in our representation. A subsequence of a melody is a theme if it:

is at least five notes long
appears (potentially transposed -- see below) again somewhere else in the piece of music
is disjoint from (i.e., non-overlapping with) at least one of its other appearance(s)

Transposed means that a constant positive or negative value is added to every note value in the theme subsequence.

Given a melody, compute the length (number of notes) of the longest theme.

One second time limit for this problem’s solutions!

Input

The input contains several test cases. The first line of each test case contains the integer N. The following n integers represent the sequence of notes.

The last test case is followed by one zero.

Output

For each test case, the output file should contain a single line with a single integer that represents the length of the longest theme. If there are no themes, output 0.

Sample Input

30

25 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 18

82 78 74 70 66 67 64 60 65 80

0

Sample Output

5

Hint

Use scanf instead of cin to reduce the read time.

Source

LouTiancheng@POJ

求最长不可重叠子串。能够后缀数组+二分解决

先把输入的数字前后两两做差,然后建立后缀数组。二分就可以

/*************************************************************************
    > File Name: POJ1743.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com
    > Created Time: 2015年03月31日 星期二 15时43分29秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

class SuffixArray
{
    public:
        static const int N = 20010;
        int init[N];
        int X[N];
        int Y[N];
        int Rank[N];
        int sa[N];
        int height[N];
        int buc[N];
        int size;

        void clear()
        {
            size = 0;
        }

        void insert(int n)
        {
            init[size++] = n;
        }

        bool cmp(int *r, int a, int b, int l)
        {
            return (r[a] == r[b] && r[a + l] == r[b + l]);
        }

        void getsa(int m = 256)
        {
            init[size] = 0;
            int l, p, *x = X, *y = Y, n = size + 1;
            for (int i = 0; i < m; ++i)
            {
                buc[i] = 0;
            }
            for (int i = 0; i < n; ++i)
            {
                buc[x[i] = init[i]]++;
            }
            for (int i = 1; i < m; ++i)
            {
                buc[i] += buc[i - 1];
            }
            for (int i = n - 1; i >= 0; --i)
            {
                sa[--buc[x[i]]] = i;
            }
            for (l = 1, p = 1; l <= n; m = p, l *= 2)
            {
                p = 0;
                for (int i = n - l; i < n; ++i)
                {
                    y[p++] = i;
                }
                for (int i = 0; i < n; ++i)
                {
                    if (sa[i] >= l)
                    {
                        y[p++] = sa[i] - l;
                    }
                }
                for (int i = 0; i < m; ++i)
                {
                    buc[i] = 0;
                }
                for (int i = 0; i < n; ++i)
                {
                    ++buc[x[y[i]]];
                }
                for (int i = 1; i < m; ++i)
                {
                    buc[i] += buc[i - 1];
                }
                for (int i = n - 1; i >= 0; --i)
                {
                    sa[--buc[x[y[i]]]] = y[i];
                }
                int i;
                for (swap(x, y), x[sa[0]] = 0, p = 1, i = 1; i < n; ++i)
                {
                    x[sa[i]] = cmp(y, sa[i - 1], sa[i], l) ?

p - 1 : p++;
                }
                if (p >= n)
                {
                    break;
                }
            }
        }

        void getheight()
        {
            int h = 0;
            for (int i = 0; i <= size; ++i)
            {
                Rank[sa[i]] = i;
            }
            height[0] = 0;
            for (int i = 0; i < size; ++i)
            {
                if (h > 0)
                {
                    --h;
                }
                int j = sa[Rank[i] - 1];
                for (; i + h < size && j + h < size && init[i + h] == init[j + h]; ++h);
                height[Rank[i] - 1] = h;
            }
        }
        bool judge(int k)
        {
            int maxs = sa[1], mins = sa[1];
            for (int i = 1; i < size; ++i)
            {
                if (height[i] < k)
                {
                    maxs = mins = sa[i + 1];
                }
                else
                {
                    maxs = max(maxs, sa[i + 1]);
                    mins = min(mins, sa[i + 1]);
                    if (maxs - mins > k)
                    {
                        return 1;
                    }
                }
            }
            return 0;
        }

        void solve()
        {
            int l = 1, r = size;
            int mid;
            int ans = 0;
            while (l <= r)
            {
                int mid = (l + r) >> 1;
                if (judge(mid))
                {
                    l = mid + 1;
                    ans = mid;
                }
                else
                {
                    r = mid - 1;
                }
            }
            ++ans;
            printf("%d\n", ans >= 5 ? ans : 0);
        }
}SA;

int val[20010];

int main()
{
    int n;
    while (~scanf("%d", &n), n)
    {
        SA.clear();
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &val[i]);
        }
        for (int i = n; i >= 2; --i)
        {
            val[i] = val[i] - val[i - 1] + 90;
        }
        for (int i = 2; i <= n; ++i)
        {
            SA.insert(val[i]);
        }
        SA.getsa();
        SA.getheight();
        SA.solve();
    }
    return 0;
}

$(function () {
$('pre.prettyprint code').each(function () {
var lines = $(this).text().split('\n').length;
var $numbering = $('

    ').addClass('pre-numbering').hide();
    $(this).addClass('has-numbering').parent().append($numbering);
    for (i = 1; i ').text(i));
    };
    $numbering.fadeIn(1700);
    });
    });

    版权声明:本文博客原创文章。博客,未经同意,不得转载。

POJ1743---Musical Theme(+后缀数组二分法)的更多相关文章

  1. poj 1743 Musical Theme (后缀数组+二分法)

    Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16162   Accepted: 5577 De ...

  2. POJ1743 Musical Theme —— 后缀数组 重复出现且不重叠的最长子串

    题目链接:https://vjudge.net/problem/POJ-1743 Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Tot ...

  3. POJ1743 Musical Theme [后缀数组]

    Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 27539   Accepted: 9290 De ...

  4. POJ1743 Musical Theme [后缀数组+分组/并查集]

    Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 27539   Accepted: 9290 De ...

  5. POJ1743 Musical Theme(后缀数组 二分)

    Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 33462   Accepted: 11124 Description A m ...

  6. POJ-1743 Musical Theme(后缀数组)

    题目大意:给一个整数序列,找出最长的连续变化相同的.至少出现两次并且不相重叠一个子序列. 题目分析:二分枚举长度进行判定. 代码如下: # include<iostream> # incl ...

  7. poj1743 Musical Theme 后缀数组的应用(求最长不重叠重复子串)

    题目链接:http://poj.org/problem?id=1743 题目理解起来比较有困难,其实就是求最长有N(1 <= N <=20000)个音符的序列来表示一首乐曲,每个音符都是1 ...

  8. [Poj1743] [后缀数组论文例题] Musical Theme [后缀数组不可重叠最长重复子串]

    利用后缀数组,先对读入整数处理str[i]=str[i+1]-str[i]+90这样可以避免负数,计算Height数组,二分答案,如果某处H<lim则将H数组分开,最终分成若干块,判断每块中是否 ...

  9. POJ 1743 Musical Theme 后缀数组 最长重复不相交子串

    Musical ThemeTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://poj.org/problem?id=1743 Description ...

  10. Poj 1743 Musical Theme (后缀数组+二分)

    题目链接: Poj  1743 Musical Theme 题目描述: 给出一串数字(数字区间在[1,88]),要在这串数字中找出一个主题,满足: 1:主题长度大于等于5. 2:主题在文本串中重复出现 ...

随机推荐

  1. Coreseek Windows下安装调试

    由于项目需要全文检索,后面就去网上查了下资料,找到了Sphinx[中文是狮身人面像]这个全文检索引擎,听说挺好用的,不过没有中文分词.后面又去找了一下,找到了Coreseek,一款中文全文检索/搜索软 ...

  2. ————weak 和————block

    Blocks理解: Blocks可以访问局部变量,但是不能修改 如果修改局部变量,需要加__block __block int multiplier = 7; int (^myBlock)(int) ...

  3. 使用VS Code开发ASP.NET Core 应用程序

    最新教程:http://www.cnblogs.com/linezero/p/VSCodeASPNETCore.html 使用VS Code开发ASP.NET Core 应用程序 准备 1.安装VS ...

  4. vm centos 添加网卡 无配置文件

    vm centos 添加网卡 无配置文件 解决办法 [root@test ~]# ifconfig eth0 Link encap:Ethernet HWaddr 00:0C:29:C8:41:FB ...

  5. 什么是侧翼区(flanking region)和侧翼区单核苷酸多态性(Flanking SNPs)

    侧翼区(flanking region) 根据维基定义:The 5' flanking region is a region of DNA that is adjacent to the 5' end ...

  6. 自己动手搭建 Redis 环境,并建立一个 .NET HelloWorld 程序测试(转)

    关于 Redis ,下面来自百度百科: redis是一个key-value存储系统.和Memcached类似,它支持存储的value类型相对更多,包括string(字符串).list(链表).set( ...

  7. hdoj 2802 F(N)【递推 规律】

    F(N) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submis ...

  8. Selenium2(java)定位页面元素 二

    辅助工具: chrome浏览器,F12打开控制台; Firefox浏览器,F12打开控制台; 或者选中要定位的元素右键 安装firefox扩展firebug和firepath; 安装之后F12可调用f ...

  9. java拦截器(interceptor)

    1.声明式 (1)注解,使用Aspect的@Aspect (2)实现HandlerInterceptor /** * 拦截请求 * * @author Administrator * */ @Comp ...

  10. C++ 变量的引用 &amp;

    创建变量的引用:int &a = b; 引用变量a是变量b的别名:是传址操作,把变量b的数据地址赋值给变量a,a和b指向同一个数据 主要用途:用作函数的形参,通过将引用变量用作参数,函数将使用 ...