FZU-2075 Substring（后缀数组）

Description

Given a string, find a substring of it which the original string contains exactly n such substrings.

Input

There are several cases. The first line of each case contains an integer n.The second line contains a string, no longer than 100000.

Output

If the such substring doesn't exist, output "impossible", else output the substring that appeared n times in the original string.If there are multiple solutions, output lexicographic smallest substring.

Sample Input

2
ababba

Sample Output

ab


题目大意：给一个字符串，找出其恰好出现n次的字典序最小的子串。
题目分析：将所有的子串排序后，定义n块为相邻的n个子串构成的字符串集合，如果某个n块的lca大于包含它的n+1块的lca，那么这个n块的lca便是恰好出现了n次的子串。

代码如下：

//# define AC

# ifndef AC

# include<iostream>
# include<cstdio>
# include<cstring>
# include<vector>
# include<queue>
# include<list>
# include<cmath>
# include<set>
# include<map>
# include<string>
# include<cstdlib>
# include<algorithm>
using namespace std;

const int N=100000;

int SA[N+5];
int tSA[N+5];
int rk[N+5];
int cnt[N+5];
int height[N+5];
int *x,*y;

bool same(int i,int j,int k,int n)
{
    if(y[i]!=y[j]) return false;
    if(i+k<n&&j+k>=n) return false;
    if(i+k>=n&&j+k<n) return false;
    return y[i+k]==y[j+k];
}

void buildSA(char* s)
{
    x=rk,y=tSA;
    int m=130;
    int n=strlen(s);
    for(int i=0;i<m;++i) cnt[i]=0;
    for(int i=0;i<n;++i) ++cnt[x[i]=s[i]];
    for(int i=1;i<m;++i) cnt[i]+=cnt[i-1];
    for(int i=n-1;i>=0;--i) SA[--cnt[x[i]]]=i;

    for(int k=1;k<=n;k<<=1){
        int p=0;
        for(int i=n-k;i<n;++i) y[p++]=i;
        for(int i=0;i<n;++i) if(SA[i]>=k) y[p++]=SA[i]-k;

        for(int i=0;i<m;++i) cnt[i]=0;
        for(int i=0;i<n;++i) ++cnt[x[y[i]]];
        for(int i=1;i<m;++i) cnt[i]+=cnt[i-1];
        for(int i=n-1;i>=0;--i) SA[--cnt[x[y[i]]]]=y[i];

        p=1;
        swap(x,y);
        x[SA[0]]=0;
        for(int i=1;i<n;++i)
            x[SA[i]]=same(SA[i],SA[i-1],k,n)?p-1:p++;
        if(p>=n) break;
        m=p;
    }
}

void getHeight(char* s)
{
    int n=strlen(s);
    int k=0;
    for(int i=0;i<n;++i) rk[SA[i]]=i;
    for(int i=0;i<n;++i){
        if(rk[i]==0)
            height[rk[i]]=k=0;
        else{
            if(k) --k;
            int j=SA[rk[i]-1];
            while(s[i+k]==s[j+k]) ++k;
            height[rk[i]]=k;
        }
    }
}

char s[N+5];
int st[N+5][20];

void spare_table()
{
    int n=strlen(s);
    for(int i=1;i<n;++i)
        st[i][0]=height[i];
    for(int k=1;(1<<k)<=n;++k){
        for(int i=1;i+(1<<k)-1<n;++i){
            st[i][k]=min(st[i][k-1],st[i+(1<<(k-1))][k-1]);
        }
    }
}

int getST(int l,int r)
{
    int k=0;
    while((1<<(k+1))<=r-l+1) ++k;
    return min(st[l][k],st[r-(1<<k)+1][k]);
}

string solve(int m)
{
    string res="";
    int n=strlen(s);
    int a,b,c;
    for(int i=0;i+m-1<n;++i){
        a=b=c=0;
        if(m==1) a=n-SA[i];
        else a=getST(i+1,i+m-1);
        if(i+m<n) b=getST(i+1,i+m);
        if(i>0) c=getST(i,i+m-1);
        if(a>b&&a>c){
            for(int j=SA[i];j<SA[i]+a;++j) res+=s[j];
            return res;
        }
    }
    return "impossible";
}

int main()
{
    int m;
    while(~scanf("%d",&m))
    {
        scanf("%s",s);
        buildSA(s);
        getHeight(s);
        spare_table();
        cout<<solve(m)<<endl;
    }
    return 0;
}

# endif