POJ 3581 Sequence(后缀数组)
2017-03-15 20:22
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原题链接
Problem Description
Given a sequence, {A1, A2, …, An} which is guaranteed A1 > A2, …, An, you are to cut it into three sub-sequences and reverse them separately to form a new one which is the smallest possible sequence in alphabet order.The alphabet order is defined as follows: for two sequence {A1, A2, …, An} and {B1, B2, …, Bn}, we say {A1, A2, …, An} is smaller than {B1, B2, …, Bn} if and only if there exists such i ( 1 ≤ i ≤ n) so that we have Ai < Bi and Aj = Bj for each j < i.
Input
The first line contains n. (n ≤ 200000)The following n lines contain the sequence.
Output
output n lines which is the smallest possible sequence obtained.Sample Input
510
1
2
3
4
Sample Output
110
2
4
3
题目大意
给出一列数字,要求你将这列数分成三段,然后将三段分别翻转,要求使得翻转后的数列字典序最小。解题思路
首先将整个数组翻转,构建第一次后缀数组,由于有保证第一个数最大这个条件,所以可以直接根据后缀数组来确定第一个分割点。(因为前面不可能出现与此有公共前缀的子串,但是确定第二个分割点就没有那么直接因为没有这样的性质)确定第二个分割点需要将剩余数列复制一遍粘贴在后部,求第二次后缀数组,只需要保证sa的值在前一半即可。找到断点后按序输出。AC代码
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cctype> #include<algorithm> #include<cmath> #include<vector> #include<string> #include<queue> #include<list> #include<stack> #include<set> #include<map> #define ll long long #define ull unsigned long long #define rep(i,n) for(int i = 0;i < n; i++) #define fil(a,b) memset((a),(b),sizeof(a)) #define cl(a) fil(a,0) #define pb push_back #define mp make_pair #define ee 2.7182818 #define PI 3.141592653589793 #define inf 0x3f3f3f3f #define fi first #define se second #define eps 1e-7 #define mod 1000000007ll #define maxn 200100 using namespace std; int n, k; int a[maxn]; int b[maxn]; int rk[maxn], temp[maxn],sa[maxn]; int rrank[maxn+1]; int lcp[maxn]; bool compare_sa(int i, int j){ if(rk[i] != rk[j]) return rk[i] < rk[j]; else { int ri = i+k<=n?rk[i+k]:-1; int rj = j+k<=n?rk[j+k]:-1; return ri < rj; } } void construct_sa(int a[], int n, int sa[]){ for(int i=0; i<=n; i++) { sa[i] = i; rk[i] = i<n?a[i]:-1; } for(k=1; k<=n; k*=2){ sort(sa, sa+n+1, compare_sa); temp[sa[0]] = 0; for(int i=1; i<=n; i++){ temp[sa[i]] = temp[sa[i-1]] + (compare_sa(sa[i-1], sa[i])?1:0); } for(int i=0; i<=n; i++) rk[i] = temp[i]; } } void construct_lcp(int a[],int n,int *sa,int *lcp) { for(int i=0;i<=n;++i) { rrank[sa[i]]=i; } int h=0; lcp[0]=0; for(int i=0;i<n;++i) { int j=sa[rrank[i]-1]; if(h>0) h--; for(;j+h<n&&i+h<n;h++) { if(a[j+h]!=a[i+h]) break; } lcp[rrank[i]-1]=h; } } int main(void) { cin>>n; for(int i=0;i<n;++i) scanf("%d",&b[i]); reverse_copy(b,b+n,a); construct_sa(a,n,sa); int p1; for(int i=0;i<n;++i) { p1=n-sa[i]; if(p1>=1&&n-p1>=2) break; } int m=n-p1; reverse_copy(b+p1,b+n,a); reverse_copy(b+p1,b+n,a+m); construct_sa(a,m*2,sa); int p2; for(int i=0;i<=2*m;++i) { p2=p1+m-sa[i]; if(p2-p1>=1&&n-p2>=1) break; } reverse(b,b+p1); reverse(b+p1,b+p2); reverse(b+p2,b+n); for(int i=0;i<n;++i) printf("%d\n",b[i]); return 0; }
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