zoukankan      html  css  js  c++  java
  • poj 1743

    Musical Theme
    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 24835   Accepted: 8377

    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

    SA

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #define MN 20003
    using namespace std;
    
    int n;
    char s1[MN];
    int s[MN],a[MN];
    int v[MN],sa[MN],q[MN],rank[MN],h[MN],mmh=0,len;
    inline void gr(int x){
        rank[sa[1]]=1;
        for (int i=2;i<=n;i++) rank[sa[i]]=(s[sa[i]]==s[sa[i-1]]&&s[sa[i]+x]==s[sa[i-1]+x])?rank[sa[i-1]]:rank[sa[i-1]]+1;
        for (int i=1;i<=n;i++) s[i]=rank[i];
    }
    inline void gv(){memset(v,0,sizeof(v));for (int i=1;i<=n;i++) v[s[i]]++;for (int i=1;i<=2e4;i++)v[i]+=v[i-1];}
    inline void gsa(){
        gv();for (int i=n;i>=1;i--) sa[v[s[i]]--]=i;gr(0);
        for (int i=1;i<n;i<<=1){
            gv();for (int j=n;j>=1;j--) if (sa[j]>i) q[v[s[sa[j]-i]]--]=sa[j]-i;
            for (int j=n-i+1;j<=n;j++) q[v[s[j]]--]=j;
            for (int j=1;j<=n;j++) sa[j]=q[j];gr(i);
            if (rank[sa[n]]==n) return;
        }
    }
    inline void gh(){for (int i=1,k=0,j;i<=n;h[rank[i++]]=k) for (k?k--:0,j=sa[rank[i]-1];a[i+k]==a[j+k]&&i+k<=n&&j+k<=n;k++);}
    int main(){
        scanf("%d",&n);
        while(n){
            for (int i=1;i<=n;i++) scanf("%d",&a[i]);
            if(n<10){printf("0
    ");scanf("%d",&n);continue;}
            n--;
            for (int i=1;i<=n;i++) s[i]=a[i+1]-a[i]+90;s[n+1]=0;
            for (int i=1;i<=n;i++) a[i]=s[i];
            gsa();gh();
            int l=4,r=2e4,mid,bo=1,ma,mi,i,j,k;
            while(l<r){
                mid=(l+r+1)>>1;
                for (i=1,j,k=2;i<=n;i=k++){
                    ma=0;mi=2e4;
                    while (h[k]>=mid&&k<=n) k++;
                    for (j=i;j<k;j++){
                        if (ma<sa[j]) ma=sa[j];
                        if (mi>sa[j]) mi=sa[j];
                    }
                    if (ma-mi>=mid) break;
                }
                if (i>n) r=mid-1;else l=mid;
            }
            l=l<4?0:l+1;
            printf("%d
    ",l);
            scanf("%d",&n);
        }
    }
    940K 250MS G++ 1716B
     
  • 相关阅读:
    js浅拷贝和深拷贝
    使用slice和concat对数组的深拷贝和浅拷贝
    JS数组常用方法---8、concat方法
    JS数组常用方法---7、join方法
    js中将类数组转换为数组的几种方法
    JS 使用const声明常量的本质(很多人都有误解)
    JS中对象数组按照对象的某个属性进行排序
    vue源码分析参考---2、数据代理
    vue源码分析参考---1、准备工作
    ES6课程---5、形参默认值
  • 原文地址:https://www.cnblogs.com/Enceladus/p/5461544.html
Copyright © 2011-2022 走看看