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

2010 ACM-ICPC Multi-University Training Contest（9）——Host by HNU

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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 }