Bzoj1640: [usaco Nov] best cow line queue Transformation

1640: [usaco Nov] best cow line queue transformation time limit: 5 sec memory limit: 64 MB
Submit: 493 solved: 253
[Submit] [Status] Description

FJ plans to bring his cute N (1 ≤ n ≤ 2,000) cows to the "best old farmer Of The Year" competition. during the competition, each farmer put his milk steak into a column and then prepared to pass the test by the judges. in the competition, the initial letter of every cow is abbreviated as the first letter. For example, FJ carries Bessie, Sylvia, and Dora, this can be abbreviated as BSD. FJ only needs to rearrange a sequence of cows and then participate in the competition. he can let the first or last cow in the sequence come out and stand at the end of the new queue. to win the championship, the powerful FJ must make the Lexicographic Order of the new queue as small as possible. give you an initial sequence of cows (indicated by a letter in the first place), then form a new sequence according to the above rules, and make the Lexicographic Order of the new sequence as small as possible.

Input

Row 1st: an integer n.

Row 2nd to row n + 1: Each line has an uppercase letter, indicating the first letter of the cow in the initial sequence.

Output

The minimum lexicographic sequence. A line break is required for each output of 80 letters!

 

Sample input6
A
C
D
B
C
B
Sample outputabcbcd
Hint

 

 

Source

Silver

Problem: The same queue is changed, but it seems that the data is small and it is greedy... The code for the previous question is now... Code:

 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #include<string>12 #define inf 100000000013 #define maxn 7000014 #define maxm 500+10015 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 using namespace std;19 inline int read()20 {21     int x=0,f=1;char ch=getchar();22     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}23     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}24     return x*f;25 }26 inline bool cmp(int *r,int a,int b,int l)27 {28     return r[a]==r[b]&&r[a+l]==r[b+l];29 }30 int wa[maxn],wb[maxn],wr[maxn],sa[maxn],rank[maxn],a[maxn],n;31 void da(int *r,int *sa,int n,int m)32 {33     int i,j,p,*x=wa,*y=wb,*t;34     for(i=0;i<m;i++)wr[i]=0;35     for(i=0;i<n;i++)wr[x[i]=r[i]]++;36     for(i=1;i<m;i++)wr[i]+=wr[i-1];37     for(i=n-1;i>=0;i--)sa[--wr[x[i]]]=i;38     for(j=1,p=1;p<n;j<<=1,m=p)39      {40         for(p=0,i=n-j;i<n;i++)y[p++]=i;41         for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;42         for(i=0;i<m;i++)wr[i]=0;43         for(i=0;i<n;i++)wr[x[y[i]]]++;44         for(i=1;i<m;i++)wr[i]+=wr[i-1];45         for(i=n-1;i>=0;i--)sa[--wr[x[y[i]]]]=y[i];46         for(t=x,x=y,y=t,p=1,i=1,x[sa[0]]=0;i<n;i++)47         x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;48      }49 }50 int main()51 {52     n=read();53     for(int i=0;i<n;i++)54      {55         char ch=‘ ‘;56         while(ch<‘A‘||ch>‘Z‘)ch=getchar();57         a[i]=int(ch-‘A‘+1);58      }59     int tot=2*n+1; 60     a[n]=0;a[tot]=0;61     for(int i=1;i<=n;i++)a[n+i]=a[n-i];62     da(a,sa,tot+1,28);63     //for(int i=0;i<=tot;i++)cout<<i<<‘ ‘<<sa[i]<<endl;64     for(int i=1;i<=tot;i++)rank[sa[i]]=i;65     //for(int i=0;i<=tot;i++)cout<<i<<‘ ‘<<rank[i]<<endl;66     int l=0,r=n+1;67     while(l+r-n-1<n)68     {69         if(rank[l]<rank[r])putchar(a[l++]+‘A‘-1);else putchar(a[r++]+‘A‘-1);70         if(!((l+r-n-1)%80))putchar(‘\n‘);71     } 72     return 0;73 }

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Bzoj1640: [usaco Nov] best cow line queue Transformation