Stammering Aliens
Time Limit: 2000MS | Memory Limit: 65536K | |
Description
Thus, the message contains possibly overlapping repetitions of the same words over and over again. As a result, Ellie turns to you, S.R. Hadden, for help in identifying the gist of the message.
Given an integer m, and a string s, representing the message, your task is to find the longest substring of s that appears at least m times. For example, in the message baaaababababbababbab, the length-5 word babab is contained 3 times, namely at positions 5, 7 and 12 (where indices start at zero). No substring appearing 3 or more times is longer (see the first example from the sample input). On the other hand, no substring appears 11 times or more (see example 2). In case there are several solutions, the substring with the rightmost occurrence is preferred (see example 3).
Input
Output
Sample Input
3
baaaababababbababbab
11
baaaababababbababbab
3
cccccc
0
Sample Output
5 12
none
4 2
分析:求出现至少m次的最长字符子串及最大位置;
后缀数组+set;(二分hash)
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
#define freopen freopen("in.txt","r",stdin)
const int maxn=4e4+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
inline ll read()
{
ll x=;int f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,k,t,cntA[maxn],cntB[maxn],sa[maxn],lev[maxn],height[maxn],A[maxn],B[maxn],tsa[maxn];
char ch[maxn];
void solve()
{
for (int i = ; i < ; i ++) cntA[i] = ;
for (int i = ; i <= n; i ++) cntA[ch[i]] ++;
for (int i = ; i < ; i ++) cntA[i] += cntA[i - ];
for (int i = n; i; i --) sa[cntA[ch[i]] --] = i;
lev[sa[]] = ;
for (int i = ; i <= n; i ++)
{
lev[sa[i]] = lev[sa[i - ]];
if (ch[sa[i]] != ch[sa[i - ]]) lev[sa[i]] ++;
}
for (int l = ; lev[sa[n]] < n; l <<= )
{
for (int i = ; i <= n; i ++) cntA[i] = ;
for (int i = ; i <= n; i ++) cntB[i] = ;
for (int i = ; i <= n; i ++)
{
cntA[A[i] = lev[i]] ++;
cntB[B[i] = (i + l <= n) ? lev[i + l] : ] ++;
}
for (int i = ; i <= n; i ++) cntB[i] += cntB[i - ];
for (int i = n; i; i --) tsa[cntB[B[i]] --] = i;
for (int i = ; i <= n; i ++) cntA[i] += cntA[i - ];
for (int i = n; i; i --) sa[cntA[A[tsa[i]]] --] = tsa[i];
lev[sa[]] = ;
for (int i = ; i <= n; i ++)
{
lev[sa[i]] = lev[sa[i - ]];
if (A[sa[i]] != A[sa[i - ]] || B[sa[i]] != B[sa[i - ]]) lev[sa[i]] ++;
}
}
for (int i = , j = ; i <= n; i ++)
{
if (j) j --;
while (ch[i + j] == ch[sa[lev[i] - ] + j]) j ++;
height[lev[i]] = j;
}
}
multiset<int>p;
set<int>q;
int main()
{
int i,j;
while(~scanf("%d",&m)&&m)
{
scanf("%s",ch+);
n=strlen(ch+);
if(m==)
{
printf("%d %d\n",n,);
continue;
}
solve();
p.clear(),q.clear();
rep(i,,m-)p.insert(height[i]),q.insert(sa[i]);
int ans=,pos=-;
rep(i,m,n)
{
p.erase(p.lower_bound(height[i-m+]));
p.insert(height[i]);
q.insert(sa[i]);
if(ans<=*p.begin())
{
auto x=q.end();
x--;
if(ans<*p.begin()||*x->pos)pos=*x-;
ans=*p.begin();
}
q.erase(sa[i-m+]);
}
if(ans)printf("%d %d\n",ans,pos);
else puts("none");
}
//system("Pause");
return ;
}
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