hdu 5030 Rabbit's String(后缀数组&二分法)

Rabbit's String

Time Limit: 40000/20000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 288    Accepted Submission(s): 108

Problem Description

Long long ago, there lived a lot of rabbits in the forest. One day, the king of the rabbit kingdom got a mysterious string and he wanted to study this string.



At first, he would divide this string into no more than k substrings. Then for each substring S, he looked at all substrings of S, and selected the one which has the largest dictionary order. Among those substrings selected in the second round, the king then
choose one which has the largest dictionary order, and name it as a "magic string".



Now he wanted to figure out how to divide the string so that the dictionary order of that "magic string" is as small as possible.

 

Input

There are at most 36 test cases.



For each test case, the first line contains a integer k indicating the maximum number of substrings the king could divide, and the second line is the original mysterious string which consisted of only lower letters.



The length of the mysterious string is between 1 and 10⁵ and k is between 1 and the length of the mysterious string, inclusive.

  

The input ends by k = 0.

 

Output

For each test case, output the magic string.

 

Sample Input

3
bbaa
2
ababa
0

 

Sample Output

b
ba
Hint
   For the first test case, the king may divide the string into "b", "b" and "aa".
   For the second test case, the king may divide the string into "aba" and "ba".
 

 

Source

field=problem&key=2014+ACM%2FICPC+Asia+Regional+Guangzhou+Online&source=1&searchmode=source">2014 ACM/ICPC Asia Regional Guangzhou Online

 

Recommend

hujie   |   We have carefully selected several similar problems for you:  5053 5052 5051 5050 

pid=5049" target="_blank">5049 

 题意：

给你一个长度不超过1e5的串。

你最多能够把它切成k个连续的子串。然后对于切出来的子串拿出他们子串字典序最大的那个(子串的子串)。

然后把全部拿出来的子串的子串字典序最大的那个串叫魔法串。

如今要你输出字典序最小的魔法串。

思路：

最大中的最小。非常easy想到二分(如今都是条件反射了)。首先我们求出sa,height。然后就能够求出f[i]即排名前i的后缀中有多少个不同的子串。然后我们二分魔法串的排名。通过排名我们能够借助f数组定位魔法串所属后缀的排名pos和魔法串的长度len。如今要解决的是怎么推断方案是否可行了。对于排名大于pos的后缀x假设tp=lcp(pos,x)=0肯定无解。无论怎么切都没用。否则我们能够在[sa[x],sa[x]+tp-1]选个位置切一刀。对于pos后的后缀都标记一次。然后贪心来切。看最少切多少刀。然后就能够判定了。

具体见代码：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
typedef long long ll;
char txt[maxn];
int sa[maxn],T1[maxn],T2[maxn],ct[maxn],he[maxn],rk[maxn],n,m,cut;
int mk[maxn];
ll f[maxn],ans;
void getsa(char *st)
{
    int i,k,p,*x=T1,*y=T2;
    for(i=0; i<m; i++) ct[i]=0;
    for(i=0; i<n; i++) ct[x[i]=st[i]]++;
    for(i=1; i<m; i++) ct[i]+=ct[i-1];
    for(i=n-1; i>=0; i--)
        sa[--ct[x[i]]]=i;
    for(k=1,p=1; p<n; k<<=1,m=p)
    {
        for(p=0,i=n-k; i<n; i++) y[p++]=i;
        for(i=0; i<n; i++) if(sa[i]>=k) y[p++]=sa[i]-k;
        for(i=0; i<m; i++) ct[i]=0;
        for(i=0; i<n; i++) ct[x[y[i]]]++;
        for(i=1; i<m; i++) ct[i]+=ct[i-1];
        for(i=n-1; i>=0; i--) sa[--ct[x[y[i]]]]=y[i];
        for(swap(x,y),p=1,x[sa[0]]=0,i=1; i<n; i++)
            x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
    }
}
void gethe(char *st)
{
    int i,j,k=0;
    for(i=0;i<n;i++) rk[sa[i]]=i;
    for(i=0;i<n-1;i++)
    {
        if(k) k--;
        j=sa[rk[i]-1];
        while(st[i+k]==st[j+k]) k++;
        he[rk[i]]=k;
    }
}
bool isok(ll p)
{
    int pos,len,i,pp,cnt;
    pos=lower_bound(f+1,f+1+n,p)-f;//定位sa
    len=he[pos]+p-f[pos-1];//确定串长
    for(i=0;i<n;i++)
        mk[i]=-1;
    if(n-sa[pos]>len)//看自己所属后缀是否要切
        mk[sa[pos]]=sa[pos]+len-1;
    for(i=pos+1;i<=n;i++)
    {
        if(he[i]==0)
            return false;
        len=min(len,he[i]);//lcp
        mk[sa[i]]=sa[i]+len-1;//排序比pos大一定要切割。
 
}
    pp=n,cnt=0;
    for(i=0;i<n;i++)
    {
        if(mk[i]!=-1)//能不切先不切和后面的一起切。贪心的思想。
            pp=min(pp,mk[i]);
        if(pp==i)
        {
            cnt++;
            if(cnt>cut)
                return false;
            pp=n;
        }
    }
    return cnt<cut;//切cnt次就是cnt+1块。
 
}
int main()
{
    int i,pos,len;
    ll low,hi,mid;
 
    while(scanf("%d",&cut),cut)
    {
        scanf("%s",txt);
        n=strlen(txt)+1;
        m=128;
        getsa(txt);
        gethe(txt);
        n--;
        f[1]=n-sa[1];
        for(i=2;i<=n;i++)
            f[i]=f[i-1]+n-sa[i]-he[i];
        low=1,hi=f[n],ans=1;
        while(low<=hi)
        {
            mid=(low+hi)>>1;
            if(isok(mid))
                ans=mid,hi=mid-1;
            else
                low=mid+1;
        }
        pos=lower_bound(f+1,f+1+n,ans)-f;
        len=he[pos]+ans-f[pos-1];
        txt[sa[pos]+len]=0;
        printf("%s\n",txt+sa[pos]);
    }
    return 0;
}

版权声明：本文博主原创文章，博客，未经同意不得转载。