Boring counting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

Recommend

zhengfeng   |   We have carefully selected several similar problems for you:  3517 3520 3519 3521 3522

Problem : 3518 ( Boring counting )     Judge Status : Accepted
RunId : 14564325    Language : C++    Author : lwj1994
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta

ac代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int s[2002];
char str[2002];
int sa[2002],t1[2002],t2[2002],c[2002];
int Rank[2002],height[2002],ans;
void build_sa(int s[],int n,int m)
{
    int i,j,p,*x=t1,*y=t2;
    for(i=0;i<m;i++)
        c[i]=0;
    for(i=0;i<n;i++)
        c[x[i]=s[i]]++;
    for(i=1;i<m;i++)
        c[i]+=c[i-1];
    for(i=n-1;i>=0;i--)
        sa[--c[x[i]]]=i;
    for(j=1;j<=n;j<<=1)
    {
        p=0;
        for(i=n-j;i<n;i++)
            y[p++]=i;
        for(i=0;i<n;i++)
            if(sa[i]>=j)
                y[p++]=sa[i]-j;
        for(i=0;i<m;i++)
            c[i]=0;
        for(i=0;i<n;i++)
            c[x[y[i]]]++;
        for(i=1;i<m;i++)
            c[i]+=c[i-1];
        for(i=n-1;i>=0;i--)
            sa[--c[x[y[i]]]]=y[i];
        swap(x,y);
        p=1;
        x[sa[0]]=0;
        for(i=1;i<n;i++)
            x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
        if(p>=n)
            break;
        m=p;
    }
}
void getHeight(int s[],int n)
{
    int i,j,k=0;
    for(i=0;i<=n;i++)
        Rank[sa[i]]=i;
    for(i=0;i<n;i++)
    {
        if(k)
            k--;
        j=sa[Rank[i]-1];
        while(s[i+k]==s[j+k])
            k++;
        height[Rank[i]]=k;
    }
}
int judge(int n,int len)
{
	int maxn=sa[0],minn=sa[0],ans=0;
	int i,j;
	for(i=1;i<=n;i++)
	{
		if(height[i]<len)
		{
			if(maxn-minn>=len)
				ans++;
			maxn=minn=sa[i];
		}
		else
		{
			if(maxn<sa[i])
				maxn=sa[i];
			if(minn>sa[i])
				minn=sa[i];
		}
	}
	if(maxn-minn>=len)
		ans++;
	return ans;
}
int main()
{
	int n;
	while(scanf("%s",str)!=EOF)
	{
		int i;
		if(strcmp(str,"#")==0)
			break;
		int len=strlen(str);
		for(i=0;i<len;i++)
			s[i]=str[i]-'a'+1;
		s[len]=0;
		build_sa(s,len+1,30);
		getHeight(s,len);
		int l=0,r=len;
		ans=0;
		for(i=1;i<=len/2;i++)
		{
			ans+=judge(len,i);

		}
		printf("%d\n",ans);
	}
}

版权声明：本文为博主原创文章，未经博主允许不得转载。