「hdu3518」Boring counting
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”).
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
题解
枚举子串长度T T将h分组，根据max(sa)-min(sa)判断解。。。
主要是模板练习
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 |
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<set> #include<ctime> #include<vector> #include<queue> #include<algorithm> #include<map> #include<cmath> #define inf 1000000000 #define pa pair<int,int> #define ll long long using namespace std; int n,k,p,q,ans; int a[1005],v[1005]; int sa[2][1005],rk[2][1005],h[1005]; char ch[1005]; void calsa(int *sa,int *rk,int *SA,int *RK) { for(int i=1;i<=n;i++)v[rk[sa[i]]]=i; for(int i=n;i;i--) if(sa[i]>k) SA[v[rk[sa[i]-k]]--]=sa[i]-k; for(int i=n-k+1;i<=n;i++)SA[v[rk[i]]--]=i; for(int i=1;i<=n;i++) RK[SA[i]]=RK[SA[i-1]]+(rk[SA[i-1]]!=rk[SA[i]]||rk[SA[i-1]+k]!=rk[SA[i]+k]); } void getsa() { p=0,q=1; for(int i=1;i<=n;i++)v[a[i]]++; for(int i=1;i<=30;i++)v[i]+=v[i-1]; for(int i=1;i<=n;i++)sa[p][v[a[i]]--]=i; for(int i=1;i<=n;i++) rk[p][sa[p][i]]=rk[p][sa[p][i-1]]+(a[sa[p][i-1]]!=a[sa[p][i]]); for(k=1;k<n;k<<=1,swap(p,q)) calsa(sa[p],rk[p],sa[q],rk[q]); } void geth() { k=0; for(int i=1;i<=n;i++) if(rk[p][i]==1)h[rk[p][i]]=0; else { int j=sa[p][rk[p][i]-1]; while(a[i+k]==a[j+k])k++; h[rk[p][i]]=k;if(k>0)k--; } } bool solve(int x) { bool flag=0; int mx=sa[p][1],mn=sa[p][1]; for(int i=2;i<=n;i++) if(h[i]<x) { if(mx-mn>=x)ans++,flag=1; mn=mx=sa[p][i]; } else { mn=min(mn,sa[p][i]); mx=max(mx,sa[p][i]); } if(mx-mn>=x)ans++,flag=1; return flag; } int main() { while(scanf("%s",ch+1)) { if(ch[1]=='#')break; ans=0; memset(a,0,sizeof(a)); memset(v,0,sizeof(v)); memset(rk,0,sizeof(rk)); n=strlen(ch+1); for(int i=1;i<=n;i++)a[i]=ch[i]-'a'+1; getsa(); geth(); for(int i=1;i<=(n+1)/2;i++) if(!solve(i))break; printf("%d\n",ans); } return 0; } |
Subscribe
计数器 9302184
OI 课件=>
OI 课件=>