POJ3623:Best Cow Line, Gold(后缀数组)
POJ3623:Best Cow Line, Gold(后缀数组)
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Description
FJ is about to take his N (1 ≤ N ≤ 30,000) cows to the annual”Farmer of the Year” competition. In this contest every farmer arranges his cows in a line and herds them past the judges.
The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows’ names.
FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.
FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he’s finished, FJ takes his cows for registration in this new order.
Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.
Input
* Line 1: A single integer: N * Lines 2..N+1: Line i+1 contains a single initial (‘A’..’Z’) of the cow in the ith position in the original line
Output
The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows (‘A’..’Z’) in the new line.
Sample Input
6
A
C
D
B
C
BSample Output
ABCBCDSource
#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;#define LS 2*i#define RS 2*i+1#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 2000005#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8int wa[N],wb[N],wsf[N],wv[N],sa[N];int rank[N],height[N],s[N];//sa:字典序中排第i位的起始位置在str中第sa[i]//rank:就是str第i个位置的后缀是在字典序排第几//height：字典序排i和i-1的后缀的最长公共前缀int cmp(int *r,int a,int b,int k){ return r[a]==r[b]&&r[a+k]==r[b+k];}void getsa(int *r,int *sa,int n,int m)//n要包括末尾加入的0{ int i,j,p,*x=wa,*y=wb,*t; for(i=0; i<m; i++) wsf[i]=0; for(i=0; i<n; i++) wsf[x[i]=r[i]]++; for(i=1; i<m; i++) wsf[i]+=wsf[i-1]; for(i=n-1; i>=0; i--) sa[--wsf[x[i]]]=i; p=1; j=1; for(; p<n; j*=2,m=p) { for(p=0,i=n-j; i<n; i++) y[p++]=i; for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j; for(i=0; i<n; i++) wv[i]=x[y[i]]; for(i=0; i<m; i++) wsf[i]=0; for(i=0; i<n; i++) wsf[wv[i]]++; for(i=1; i<m; i++) wsf[i]+=wsf[i-1]; for(i=n-1; i>=0; i--) sa[--wsf[wv[i]]]=y[i]; t=x; x=y; y=t; x[sa[0]]=0; for(p=1,i=1; i<n; i++) x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++; }}void getheight(int *r,int n)//n不保存最后的0{ int i,j,k=0; for(i=1; i<=n; i++) rank[sa[i]]=i; for(i=0; i<n; i++) { if(k) k--; else k=0; j=sa[rank[i]-1]; while(r[i+k]==r[j+k]) k++; height[rank[i]]=k; }}char str[10];int main(){ int n,i,j,k; while(~scanf("%d",&n)) { int len = 0; for(i = 0; i<n; i++) { scanf("%s",str); s[i] = str[0]; s[2*n-i] = str[0]; } s[n] = 0; s[2*n+2] = 0; getsa(s,sa,2*n+2,200); for(i = 0; i<2*n+2; i++) rank[sa[i]] = i; int l = 0,r = n+1; while((len++)<n) { if(rank[l]<rank[r]) { printf("%c",s[l]); l++; } else { printf("%c",s[r]); r++; } if(len%80==0) printf("\n"); } if(len%80) printf("\n"); } return 0;}发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/116913.html原文链接：https://javaforall.cn
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