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 10
5 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 2014 ACM/ICPC Asia Regional Guangzhou Online
Recommend hujie | We have carefully selected several similar problems for you: 5053 5052 5051 5050 5049 题意： 给你一个长度不超过1e5的串。你最多可以把它切成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)。然后把所有拿出来的子串的子串字典序最大的那个串叫魔法串。现在要你输出字典序最小的魔法串。 思路： 最大中的最小。很容易想到二分(现在都是条件反射了)。首先我们求出sa,height。然后就可以求出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。通过排名我们可以借助f数组定位魔法串所属后缀的排名pos和魔法串的长度len。现在要解决的是怎么判断方案是否可行了。对于排名大于pos的后缀x如果tp=lcp(pos,x)=0肯定无解。不管怎么切都没用。否则我们可以在[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。然后就可以判定了。 详细见代码：
#include
#include
#include
#include
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
=0; i--) sa[--ct[x[i]]]=i; for(k=1,p=1; p
=k) y[p++]=sa[i]-k; for(i=0; i
=0; i--) sa[--ct[x[y[i]]]]=y[i]; for(swap(x,y),p=1,x[sa[0]]=0,i=1; i
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大一定要分割。 } pp=n,cnt=0; for(i=0;i
cut) return false; pp=n; } } return cnt
>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; }