zoukankan      html  css  js  c++  java
  • HDU3518 Boring counting

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3182    Accepted Submission(s): 1319


    Problem Description
    035 now faced a tough problem,his english teacher gives him a string,which consists with n lower case letter,he must figure out how many substrings appear at least twice,moreover,such apearances can not overlap each other.
    Take aaaa as an example.”a” apears four times,”aa” apears two times without overlaping.however,aaa can’t apear more than one time without overlaping.since we can get “aaa” from [0-2](The position of string begins with 0) and [1-3]. But the interval [0-2] and [1-3] overlaps each other.So “aaa” can not take into account.Therefore,the answer is 2(“a”,and “aa”).
     
    Input
    The input data consist with several test cases.The input ends with a line “#”.each test case contain a string consists with lower letter,the length n won’t exceed 1000(n <= 1000).
     
    Output
    For each test case output an integer ans,which represent the answer for the test case.you’d better use int64 to avoid unnecessary trouble.
     
    Sample Input
    aaaa ababcabb aaaaaa #
     
    Sample Output
    2 3 3
     
    Source
     
    Recommend
    zhengfeng

    字符串 后缀数组

    求至少不重叠出现2次以上的子串有多少个

    建立后缀数组,按height把后缀们分成不同的组,同一组内都是相同的后缀,统计它们出现的最大最小位置即可

     1 /*by SilverN*/
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 #include<vector>
     8 using namespace std;
     9 const int mxn=100010;
    10 int read(){
    11     int x=0,f=1;char ch=getchar();
    12     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    13     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    14     return x*f;
    15 }
    16 int sa[mxn],rk[mxn],ht[mxn];
    17 int wa[mxn],wb[mxn],wv[mxn],cnt[mxn];
    18 char s[mxn];
    19 inline int cmp(int *r,int a,int b,int l){
    20     return r[a]==r[b] && r[a+l]==r[b+l];
    21 }
    22 void GSA(int *sa,int n,int m){
    23     int i,j,k;
    24     int *x=wa,*y=wb;
    25     for(i=0;i<m;i++)cnt[i]=0;
    26     for(i=0;i<n;i++)cnt[x[i]=s[i]-'a'+1]++;
    27     for(i=1;i<m;i++)cnt[i]+=cnt[i-1];
    28     for(i=n-1;i>=0;i--)sa[--cnt[x[i]]]=i;
    29     for(int p=0,j=1;p<n;j<<=1,m=p){
    30         for(p=0,i=n-j;i<n;i++)y[p++]=i;
    31         for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
    32         for(i=0;i<n;i++)
    33             wv[i]=x[y[i]];
    34         for(i=0;i<m;i++)cnt[i]=0;
    35         for(i=0;i<n;i++)cnt[wv[i]]++;
    36         for(i=1;i<m;i++)cnt[i]+=cnt[i-1];
    37         for(i=n-1;i>=0;i--)sa[--cnt[wv[i]]]=y[i];
    38         swap(x,y);
    39         p=1;x[sa[0]]=0;
    40         for(i=1;i<n;i++)
    41             x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;
    42     }
    43     return;
    44 }
    45 void GHT(int n){
    46     int i,j,k=0;
    47     for(i=1;i<=n;i++)rk[sa[i]]=i;
    48     for(i=0;i<n;i++){
    49         if(k)k--;
    50         j=sa[rk[i]-1];
    51         while(s[i+k]==s[j+k])k++;
    52         ht[rk[i]]=k;
    53     }
    54     return;
    55 }
    56 int ans=0;
    57 bool solve(int n,int lim){
    58     bool flag=0;
    59     int mxpos=-100,mnpos=mxn;
    60     for(int i=1;i<=n;i++){
    61         if(ht[i]<lim){
    62             if(mxpos-mnpos>=lim){        
    63                 ans++;flag=1;
    64             }
    65             mxpos=-100,mnpos=mxn;
    66         }
    67         mxpos=max(mxpos,sa[i]);
    68         mnpos=min(mnpos,sa[i]);
    69     }
    70     if(mxpos-mnpos>=lim)flag=1,ans++;
    71     return flag;
    72 }
    73 int main(){
    74     int i,j;
    75     while(scanf("%s",s) && s[0]!='#'){
    76         int len=strlen(s);
    77         s[len]='a'-1;
    78         GSA(sa,len+1,28);
    79         GHT(len);
    80         ans=0;
    81 //        for(i=0;i<=len;i++)printf("%c",s[i]);puts("");
    82 //        for(i=1;i<=len;i++)printf("%d ",sa[i]);puts("");
    83 //        for(i=1;i<=len;i++)printf("%d ",ht[i]);puts("");
    84         for(i=1;i<len;i++){
    85             if(!solve(len,i))break;
    86         }
    87         printf("%d
    ",ans);
    88     }
    89     return 0;
    90 }
  • 相关阅读:
    ThinkPhp框架分页查询和部分框架知识
    tp框架增删改
    WAMP中mysql服务突然无法启动 解决方法
    thinkphp框架 的 链接数据库和操作数据
    php 全局使用laravel的dd和dump
    给centos装图形界面 widowsx
    marquee标签的使用
    微信公众号开发入门教程
    laravel admin引入css js报错 https
    利用Croppie裁剪图片并后台保存
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/6648740.html
Copyright © 2011-2022 走看看